Explanation:
Okay, so I need to find the smallest positive integer n such that all the roots of the equation z⁴ + z² + 1 = 0 are nth roots of unity. Hmm, let me think. First, let me recall what roots of unity are. They are complex numbers that satisfy zⁿ = 1 for some positive integer n. So, if a root of the given equation is also a root of zⁿ - 1 = 0, then it's an nth root of unity. My job is to find the smallest such n.

First, maybe I should try to find all the roots of the equation z⁴ + z² + 1 = 0. Once I have them, I can figure out their orders (the smallest n for which they are nth roots of unity) and then take the least common multiple (LCM) of all those orders. That should give me the minimal n required. Let me try that approach.

So, starting with the equation z⁴ + z² + 1 = 0. Let me see if I can factor this or simplify it somehow. Maybe using substitution? Let me set w = z². Then the equation becomes w² + w + 1 = 0. That's a quadratic equation in w. Let me solve for w.

Using the quadratic formula: w = [-1 ± √(1² - 4*1*1)] / (2*1) = [-1 ± √(1 - 4)] / 2 = [-1 ± √(-3)] / 2 = (-1 ± i√3)/2. So, w = (-1 + i√3)/2 or w = (-1 - i√3)/2.

Therefore, z² = (-1 ± i√3)/2. Now, I need to find the square roots of these complex numbers to get the roots of the original equation. Let me first recall that complex numbers can be expressed in polar form, which might make taking roots easier.

Let me consider the first value of w: (-1 + i√3)/2. Let me find its modulus and argument. The modulus is sqrt[(-1/2)² + (√3/2)²] = sqrt[(1/4) + (3/4)] = sqrt[1] = 1. So modulus is 1. The argument θ is such that cosθ = -1/2 and sinθ = √3/2. That corresponds to θ = 120° or 2π/3 radians. Similarly, the other value of w is (-1 - i√3)/2, which has modulus 1 and argument -2π/3 or 4π/3 radians.

Therefore, each w is a complex number on the unit circle at angles 2π/3 and 4π/3. Now, to find z such that z² = w, we can take square roots. The square roots of a complex number in polar form are obtained by taking the square root of the modulus and halving the angle. Since modulus is 1, the modulus of z will be 1. The arguments will be half of 2π/3 and 2π/3 + 2π, and similarly for the other w.

Wait, let me be precise. Let me take the first w: angle 2π/3. The square roots will have angles (2π/3)/2 = π/3 and (2π/3 + 2π)/2 = (8π/3)/2 = 4π/3. Similarly, for the other w with angle 4π/3, the square roots will have angles (4π/3)/2 = 2π/3 and (4π/3 + 2π)/2 = (10π/3)/2 = 5π/3. So altogether, the four roots of z⁴ + z² + 1 = 0 are:

z = e^(iπ/3), e^(i4π/3), e^(i2π/3), e^(i5π/3). Wait, hold on. Let me verify that.

Alternatively, perhaps I can write them out explicitly. For the first w: (-1 + i√3)/2. Let me note that e^(i2π/3) = cos(2π/3) + i sin(2π/3) = -1/2 + i(√3/2), which is exactly the value of w. Similarly, e^(i4π/3) = cos(4π/3) + i sin(4π/3) = -1/2 - i√3/2, which is the other w. Therefore, solving z² = e^(i2π/3) gives z = e^(iπ/3) and z = e^(iπ/3 + π) = e^(i4π/3). Similarly, solving z² = e^(i4π/3) gives z = e^(i2π/3) and z = e^(i2π/3 + π) = e^(i5π/3). Therefore, the four roots are e^(iπ/3), e^(i4π/3), e^(i2π/3), e^(i5π/3).

So, these are four roots on the unit circle at angles π/3, 2π/3, 4π/3, 5π/3. Now, each of these roots is a complex number on the unit circle, so they are all roots of unity if their angles are rational multiples of 2π. Let me check. For example, π/3 is 2π/6, so if I take n=6, then π/3 = 2π/6 would be a 6th root of unity. Similarly, 2π/3 is 4π/6, which is also a 6th root of unity. Similarly, 4π/3 is 8π/6 = 4π/3, which is equivalent to -2π/3, which is still a 6th root. Similarly, 5π/3 is 10π/6 = 5π/3, which is equivalent to -π/3, so also a 6th root. Therefore, all four roots are 6th roots of unity? Wait, that would mean n=6 suffices. But let me check if they are indeed 6th roots of unity.

Let me test e^(iπ/3). Let me compute [e^(iπ/3)]^6 = e^(i6*(π/3)) = e^(i2π) = 1. Similarly, [e^(i2π/3)]^6 = e^(i4π) = 1. Similarly for the others. So yes, they are all 6th roots of unity. But wait, the original equation is z⁴ + z² + 1 = 0. Let me check if these roots satisfy that equation. Let me take z = e^(iπ/3). Then z² = e^(i2π/3), z⁴ = e^(i4π/3). So z⁴ + z² + 1 = e^(i4π/3) + e^(i2π/3) + 1. Let me compute that. Let me recall that e^(i2π/3) + e^(i4π/3) = -1. Because e^(i2π/3) + e^(i4π/3) = cos(2π/3) + i sin(2π/3) + cos(4π/3) + i sin(4π/3) = (-1/2 + i√3/2) + (-1/2 - i√3/2) = -1. Therefore, adding 1 gives -1 + 1 = 0. So yes, that works. Similarly for the other roots. Therefore, all four roots are 6th roots of unity. Therefore, n=6? But the problem is asking for the smallest n such that all roots are nth roots of unity. Is 6 the minimal such n?

Wait, but let me check if maybe they are roots of unity of smaller order. For example, is e^(iπ/3) a 3rd root of unity? Let me see. The 3rd roots of unity are e^(i0), e^(i2π/3), e^(i4π/3). So e^(iπ/3) is not one of them. Similarly, e^(iπ/3) is a primitive 6th root of unity. Let me check the order. The order of e^(iπ/3) is the smallest positive integer k such that (e^(iπ/3))^k = 1. That would require k*(π/3) = 2π*m for some integer m, so k/3 = 2m, so k = 6m. Therefore, the minimal k is 6. Therefore, e^(iπ/3) is a primitive 6th root of unity. Similarly for the others. For example, e^(i2π/3) is a primitive 3rd root of unity? Wait, let me check. The order of e^(i2π/3) is the minimal k such that (e^(i2π/3))^k = 1. That requires k*(2π/3) = 2π*m => 2k/3 = 2m => k/3 = m. Therefore, minimal k is 3. Therefore, e^(i2π/3) is a primitive 3rd root of unity. Similarly, e^(i4π/3) is also a primitive 3rd root of unity, and e^(i5π/3) is a primitive 6th root of unity. Therefore, the roots have orders 3 and 6. Therefore, the minimal n such that all roots are nth roots of unity must be the least common multiple of 3 and 6, which is 6. Therefore, n=6.

But wait, hold on. Let me check if maybe the roots are also roots of unity of a smaller order? For example, if there is an n smaller than 6 that works? Let me check n=3. If n=3, then the 3rd roots of unity are 1, e^(i2π/3), e^(i4π/3). But the roots of our equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). However, e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity. So n=3 is too small. Similarly, n=4? Let's see. The 4th roots of unity are 1, i, -1, -i. Our roots are at angles π/3, 2π/3, 4π/3, 5π/3. None of these angles are multiples of π/2 (since π/3 ≈ 60°, which is not 90°, 180°, etc.), so n=4 is too small. How about n=5? The 5th roots of unity have angles 2π/5, 4π/5, 6π/5, 8π/5, which are not equal to π/3, 2π/3, etc. So n=5 is out. n=6 seems to be the first candidate.

But maybe there's a smaller n than 6? Let me think again. Let me check n=12? No, 6 is smaller. Wait, perhaps there is a smaller n? Let me check n=6. But is there a smaller n? For example, n=6 is the LCM of 3 and 6. But maybe if all roots have orders dividing n. Since some roots have order 3 and some have order 6, then n must be a multiple of 6. Therefore, the minimal n is 6.

Wait, but let me check if maybe the roots can be expressed as roots of unity of order less than 6? For example, is there an n such that all these roots are nth roots of unity? Let me think about the minimal polynomial. The equation z⁴ + z² + 1 = 0. Let me see if this polynomial is related to cyclotomic polynomials. Let me recall that cyclotomic polynomials are minimal polynomials over integers for primitive roots of unity.

For example, Φ₃(z) = z² + z + 1, Φ₆(z) = z² - z + 1, Φ₁₂(z) = z⁴ - z² + 1, etc. Wait, let me check. The cyclotomic polynomial Φ_n(z) is the product of (z - ζ) where ζ ranges over the primitive nth roots of unity. So, perhaps our polynomial z⁴ + z² + 1 can be related to Φ₁₂(z) or something else?

Wait, let me note that z⁶ - 1 = (z³ - 1)(z³ + 1) = (z - 1)(z² + z + 1)(z + 1)(z² - z + 1). Therefore, Φ₃(z) is z² + z + 1, Φ₆(z) is z² - z + 1. Also, Φ₁₂(z) is the minimal polynomial for primitive 12th roots of unity. Let me compute Φ₁₂(z). The formula for cyclotomic polynomials is Φ_n(z) = ∏_{k=1, gcd(k,n)=1}^n (z - e^(2πik/n)). For n=12, the primitive roots are those k where gcd(k,12)=1, so k=1,5,7,11. Therefore, Φ₁₂(z) = (z - e^(iπ/6))(z - e^(i5π/6))(z - e^(i7π/6))(z - e^(i11π/6)).

Let me compute Φ₁₂(z). Alternatively, there is a formula: Φ_n(z) = ∏_{d|n} (z^{n/d} - 1)^{μ(d)} where μ is the Möbius function. But maybe it's easier to note that Φ₁₂(z) divides z^φ(12) - 1, but φ(12)=4. Wait, no, Φ_n(z) divides z^n - 1, and the product of Φ_d(z) over d|n is z^n - 1. So, for n=12, z^12 - 1 = Φ₁(z)Φ₂(z)Φ₃(z)Φ₄(z)Φ₆(z)Φ₁₂(z). Since Φ₁(z)=z-1, Φ₂(z)=z+1, Φ₃(z)=z² + z +1, Φ₄(z)=z² + 1, Φ₆(z)=z² - z +1, and Φ₁₂(z) is the remaining factor.

So, Φ₁₂(z) = (z^12 -1)/[(z -1)(z +1)(z² + z +1)(z² +1)(z² - z +1)]. Let me compute that.

First, compute (z -1)(z +1)= z² -1.

Then multiply by (z² + z +1): (z² -1)(z² + z +1) = z^4 + z³ + z² - z² - z -1 = z^4 + z³ - z -1.

Then multiply by (z² +1): (z^4 + z³ - z -1)(z² +1) = z^6 + z^5 - z³ - z + z^4 + z³ - z -1 = z^6 + z^5 + z^4 - 2z -1. Wait, maybe I need to do this more carefully.

Alternatively, perhaps there's another approach. Let me recall that Φ₁₂(z) is z^4 - z² + 1. Wait, let me check with some known values. If z^6 -1 = (z^2 -1)(z^4 + z^2 +1). Wait, no:

Wait z^6 -1 = (z^3)^2 - 1 = (z^3 -1)(z^3 +1) = (z -1)(z^2 + z +1)(z +1)(z^2 - z +1). So, z^4 + z^2 +1 is not directly a factor here. Wait, but if I take z^6 -1 divided by (z^2 -1), which is Φ₁(z)Φ₂(z), I get z^4 + z² + 1? Let me check:

(z^6 -1) / (z^2 -1) = z^4 + z² +1. Yes! Because (z^2 -1)(z^4 + z² +1) = z^6 + z^4 + z² - z^4 - z² -1 = z^6 -1. Therefore, z^4 + z² +1 = (z^6 -1)/(z² -1). Therefore, z^4 + z² +1 is equal to Φ₃(z)Φ₆(z)? Wait, because z^6 -1 = Φ₁(z)Φ₂(z)Φ₃(z)Φ₆(z)Φ₁₂(z)? Wait, earlier we had z^6 -1 = (z^3 -1)(z^3 +1) = (z -1)(z² + z +1)(z +1)(z² - z +1). Which is Φ₁(z)Φ₃(z)Φ₂(z)Φ₆(z). Therefore, Φ₁₂(z) is not a factor here. Wait, but z^6 -1 is equal to z^4 + z² +1 multiplied by z² -1. Therefore, z^4 + z² +1 is equal to (z^6 -1)/(z² -1). Therefore, if we factor z^6 -1 as (z^2 -1)(z^4 + z² +1), then z^4 + z² +1 is equal to Φ₃(z)Φ₆(z)? Because z^6 -1 factors into Φ₁(z)Φ₂(z)Φ₃(z)Φ₆(z). So, Φ₁(z)Φ₂(z)Φ₃(z)Φ₆(z) = (z^2 -1)(z^4 + z² +1). So, Φ₁(z)Φ₂(z) = z² -1, and Φ₃(z)Φ₆(z) = z^4 + z² +1. Therefore, the polynomial z^4 + z² +1 factors into Φ₃(z)Φ₆(z). Therefore, the roots of z^4 + z² +1 =0 are the roots of Φ₃(z) and Φ₆(z). But Φ₃(z) has roots the primitive 3rd roots of unity, which are e^(i2π/3) and e^(i4π/3), and Φ₆(z) has roots the primitive 6th roots of unity, which are e^(iπ/3), e^(i5π/3), e^(iπ/3), e^(i5π/3)? Wait, no, Φ₆(z) is degree φ(6)=2. The primitive 6th roots of unity are e^(iπ/3) and e^(i5π/3). Therefore, Φ₆(z)=z² - z +1, which has roots e^(iπ/3) and e^(i5π/3). Wait, let me check. If Φ₆(z) is z² - z +1, then its roots satisfy z² - z +1 =0. Solving this quadratic equation: z = [1 ± √(-3)] / 2 = [1 ± i√3]/2. Which are e^(iπ/3) and e^(i5π/3), since e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + i√3/2 and e^(i5π/3) is 1/2 - i√3/2. Therefore, indeed Φ₆(z) = z² - z +1. Wait, but our original equation is z⁴ + z² +1 =0. So if it factors into Φ₃(z)Φ₆(z), then Φ₃(z) is z² + z +1 and Φ₆(z) is z² - z +1. Multiplying them gives (z² + z +1)(z² - z +1) = z^4 - z² + z² + z² - z + z + ... Wait, let me compute it properly. (z² + z +1)(z² - z +1) = z²(z² - z +1) + z(z² - z +1) +1(z² - z +1) = z^4 - z³ + z² + z³ - z² + z + z² - z +1. Then simplifying:

z^4 + (-z³ + z³) + (z² - z² + z²) + (z - z) + 1 = z^4 + z² + 1. Yes! Therefore, z⁴ + z² +1 factors as Φ₃(z)Φ₆(z). Therefore, the roots of z⁴ + z² +1 = 0 are precisely the primitive 3rd roots of unity and the primitive 6th roots of unity.

Therefore, the roots are the primitive 3rd roots of unity (which are order 3) and the primitive 6th roots of unity (which are order 6). Therefore, to include all these roots as nth roots of unity, n needs to be a multiple of both 3 and 6, so the least common multiple of 3 and 6 is 6. Therefore, the minimal n is 6. Therefore, the answer is 6?

But let me verify once more. For example, is 6 the minimal n? Suppose someone thinks maybe n=12? But 6 is smaller. Let me check if all roots are 6th roots of unity. The primitive 3rd roots of unity are also 6th roots of unity, since 3 divides 6, so they are 6th roots. Similarly, the primitive 6th roots of unity are 6th roots. Therefore, all roots are 6th roots of unity. Therefore, n=6 works. Since there is no smaller n (as we saw for n=3, 4, 5 do not work) then the minimal n is 6.

But let me just make sure there isn't a mistake here. For example, maybe the roots are also roots of unity of a smaller order. Wait, for example, if we have a root that is a primitive 6th root, then the order is 6, so n must be multiple of 6. The primitive 3rd roots have order 3, so n must be multiple of 3. Since 6 is the LCM of 3 and 6, that's the minimal n. Therefore, yes, n=6. Therefore, I think the answer is 6.

But just to be thorough, let me check if there exists a smaller n. Let's suppose n=6 is the minimal. Let me check n=4. As before, the 4th roots of unity are 1, -1, i, -i. But our roots are at angles π/3, 2π/3, 4π/3, 5π/3. None of these angles correspond to multiples of π/2 (which would be the angles for 4th roots), so no. Similarly, n=5: 5th roots of unity have angles 2π/5, 4π/5, 6π/5, 8π/5, 10π/5=2π. None of these are π/3 or 2π/3, etc. So n=5 is out. n=6 is the first where angles π/3, 2π/3, etc., are multiples of 2π/6=π/3. Wait, 2π/6=π/3. So yes, all the roots are 6th roots of unity. So n=6.

Alternatively, maybe there is an n less than 6? For example, n=12? But 6 is smaller. Wait, the question is about the minimal n. So 6 seems to be the answer.

Wait, but let me think again. Suppose that the roots are 6th roots of unity. But maybe they are 12th roots? But since 6 is smaller, the minimal n is 6. But let me check if maybe n=12 is required? For instance, if the roots have higher order? But no, we already saw that they are order 3 and 6, so n=6 suffices.

Alternatively, maybe I made a mistake in factoring the polynomial. Let me confirm that z⁴ + z² +1 factors into Φ₃(z)Φ₆(z). As we saw, Φ₃(z) is z² + z +1 and Φ₆(z) is z² - z +1. Multiplying them gives z⁴ + z² +1, which matches. Therefore, yes, the roots are exactly the primitive 3rd and 6th roots of unity. Therefore, n must be 6.

Alternatively, maybe the question is not asking for all roots to be primitive roots of unity but just roots of unity. But even if they are not primitive, as long as they are roots of unity, then they are included in some nth roots. But since the primitive ones have orders 3 and 6, then the nth must be multiple of 6, as we said.

Therefore, I think the answer is 6. But just to make sure, let me check with specific examples. Take the root e^(iπ/3), which is a primitive 6th root of unity. So it is a 6th root of unity. The other roots: e^(i2π/3) is a primitive 3rd root of unity, so it is also a 6th root of unity. Similarly for e^(i4π/3), e^(i5π/3). Therefore, all roots are 6th roots of unity. Therefore, n=6 works. Since there is no smaller n (as n=3 fails because of e^(iπ/3), n=4 fails as angles don't match, etc.), then n=6 is indeed minimal. Therefore, the answer is 6.

**Final Answer**
The smallest positive integer $ n $ is \boxed{6}.

To solve the problem, we are given the polynomial equation:

$$
z^4 + z^2 + 1 = 0
$$

We are asked to find the **smallest positive integer $ n $** such that **all the roots** of this equation are **$ n $th roots of unity**.

---

### Step 1: Understand the nature of the roots

We begin by recognizing that a **root of unity** is a complex number $ \omega $ satisfying:

$$
\omega^n = 1
$$

So, we first identify all the roots of the given polynomial. Let's denote the roots as $ z_1, z_2, z_3, z_4 $.

Let's rewrite the equation:

$$
z^4 + z^2 + 1 = 0
$$

We make a substitution: Let $ w = z^2 $, so that the equation becomes:

$$
w^2 + w + 1 = 0
$$

This is a quadratic in $ w $, with complex solutions:

$$
w = \frac{-1 \pm i\sqrt{3}}{2}
$$

Observe that these are the **primitive 3rd roots of unity**: $ e^{2\pi i / 3} $ and $ e^{4\pi i / 3} $.

Now, since $ w = z^2 $, the roots $ z $ of the original equation are the **square roots** of the above $ w $ values. Specifically:

- $ z = \sqrt{e^{2\pi i / 3}} = e^{\pi i / 3}, e^{4\pi i / 3} $
- $ z = \sqrt{e^{4\pi i / 3}} = e^{2\pi i / 3}, e^{5\pi i / 3} $

Thus, the four roots of $ z^4 + z^2 + 1 = 0 $ are:

$$
z = e^{i\pi/3},\ e^{i2\pi/3},\ e^{i4\pi/3},\ e^{i5\pi/3}
$$

---

### Step 2: Determine which roots are roots of unity

All these roots lie on the unit circle, and their angles are rational multiples of $ \pi $. Thus, they are **roots of unity**.

But to be precise, we consider their **order** — that is, the smallest positive integer $ n $ for which $ z^n = 1 $.

Let's check each root:

- $ e^{i\pi/3} $: this is a **primitive 6th root of unity**, since $ \pi/3 = 2\pi/6 $
- $ e^{i2\pi/3} $: a **primitive 3rd root of unity**, since $ 2\pi/3 = 2\pi/3 $
- $ e^{i4\pi/3} $: same as above
- $ e^{i5\pi/3} $: same as $ e^{i\pi/3} $, also a **primitive 6th root of unity**

Therefore, the roots have **orders 3 and 6**. This means all roots are **6th roots of unity**, because the 3rd roots of unity are also 6th roots of unity (since 3 divides 6).

---

### Step 3: Find the minimal such $ n $

The problem asks for the **smallest** $ n $ such that **all** roots are $ n $th roots of unity.

We now consider the orders of the roots:
- One root has **order 3**
- Others have **order 6**

So, $ n $ must be a **multiple of both 3 and 6**, which means $ n $ must be a **multiple of the least common multiple (LCM)** of these numbers.

$$
\text{LCM}(3, 6) = 6
$$

We verify that $ n = 6 $ works:
- All roots are 6th roots of unity (since both 3 and 6 divide 6)
- $ n = 6 $ is smaller than the next candidate (e.g., $ n = 12 $, or $ n = 3 $, which does not include all roots), so it’s **minimal**

---

### Final Conclusion

All the roots of the polynomial $ z^4 + z^2 + 1 = 0 $ are 6th roots of unity, and 6 is the **smallest** positive integer for which this is true.

---

### Final Answer

$$
\boxed{6}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest positive integer n such that all the roots of the equation z⁴ + z² + 1 = 0 are nth roots of unity. Hmm, let me think. First, let me recall what roots of unity are. They are complex numbers that satisfy zⁿ = 1 for some positive integer n. So, if a root of the given equation is also a root of zⁿ - 1 = 0, then it's an nth root of unity. My job is to find the smallest such n.

First, maybe I should try to find all the roots of the equation z⁴ + z² + 1 = 0. Once I have them, I can figure out their orders (the smallest n for which they are nth roots of unity) and then take the least common multiple (LCM) of all those orders. That should give me the minimal n required. Let me try that approach.

So, starting with the equation z⁴ + z² + 1 = 0. Let me see if I can factor this or simplify it somehow. Maybe using substitution? Let me set w = z². Then the equation becomes w² + w + 1 = 0. That's a quadratic equation in w. Let me solve for w.

Using the quadratic formula: w = [-1 ± √(1² - 4*1*1)] / (2*1) = [-1 ± √(1 - 4)] / 2 = [-1 ± √(-3)] / 2 = (-1 ± i√3)/2. So, w = (-1 + i√3)/2 or w = (-1 - i√3)/2.

Therefore, z² = (-1 ± i√3)/2. Now, I need to find the square roots of these complex numbers to get the roots of the original equation. Let me first recall that complex numbers can be expressed in polar form, which might make taking roots easier.

Let me consider the first value of w: (-1 + i√3)/2. Let me find its modulus and argument. The modulus is sqrt[(-1/2)² + (√3/2)²] = sqrt[(1/4) + (3/4)] = sqrt[1] = 1. So modulus is 1. The argument θ is such that cosθ = -1/2 and sinθ = √3/2. That corresponds to θ = 120° or 2π/3 radians. Similarly, the other value of w is (-1 - i√3)/2, which has modulus 1 and argument -2π/3 or 4π/3 radians.

Therefore, each w is a complex number on the unit circle at angles 2π/3 and 4π/3. Now, to find z such that z² = w, we can take square roots. The square roots of a complex number in polar form are obtained by taking the square root of the modulus and halving the angle. Since modulus is 1, the modulus of z will be 1. The arguments will be half of 2π/3 and 2π/3 + 2π, and similarly for the other w.

Wait, let me be precise. Let me take the first w: angle 2π/3. The square roots will have angles (2π/3)/2 = π/3 and (2π/3 + 2π)/2 = (8π/3)/2 = 4π/3. Similarly, for the other w with angle 4π/3, the square roots will have angles (4π/3)/2 = 2π/3 and (4π/3 + 2π)/2 = (10π/3)/2 = 5π/3. So altogether, the four roots of z⁴ + z² + 1 = 0 are:

z = e^(iπ/3), e^(i4π/3), e^(i2π/3), e^(i5π/3). Wait, hold on. Let me verify that.

Alternatively, perhaps I can write them out explicitly. For the first w: (-1 + i√3)/2. Let me note that e^(i2π/3) = cos(2π/3) + i sin(2π/3) = -1/2 + i(√3/2), which is exactly the value of w. Similarly, e^(i4π/3) = cos(4π/3) + i sin(4π/3) = -1/2 - i√3/2, which is the other w. Therefore, solving z² = e^(i2π/3) gives z = e^(iπ/3) and z = e^(iπ/3 + π) = e^(i4π/3). Similarly, solving z² = e^(i4π/3) gives z = e^(i2π/3) and z = e^(i2π/3 + π) = e^(i5π/3). Therefore, the four roots are e^(iπ/3), e^(i4π/3), e^(i2π/3), e^(i5π/3).

So, these are four roots on the unit circle at angles π/3, 2π/3, 4π/3, 5π/3. Now, each of these roots is a complex number on the unit circle, so they are all roots of unity if their angles are rational multiples of 2π. Let me check. For example, π/3 is 2π/6, so if I take n=6, then π/3 = 2π/6 would be a 6th root of unity. Similarly, 2π/3 is 4π/6, which is also a 6th root of unity. Similarly, 4π/3 is 8π/6 = 4π/3, which is equivalent to -2π/3, which is still a 6th root. Similarly, 5π/3 is 10π/6 = 5π/3, which is equivalent to -π/3, so also a 6th root. Therefore, all four roots are 6th roots of unity? Wait, that would mean n=6 suffices. But let me check if they are indeed 6th roots of unity.

Let me test e^(iπ/3). Let me compute [e^(iπ/3)]^6 = e^(i6*(π/3)) = e^(i2π) = 1. Similarly, [e^(i2π/3)]^6 = e^(i4π) = 1. Similarly for the others. So yes, they are all 6th roots of unity. But wait, the original equation is z⁴ + z² + 1 = 0. Let me check if these roots satisfy that equation. Let me take z = e^(iπ/3). Then z² = e^(i2π/3), z⁴ = e^(i4π/3). So z⁴ + z² + 1 = e^(i4π/3) + e^(i2π/3) + 1. Let me compute that. Let me recall that e^(i2π/3) + e^(i4π/3) = -1. Because e^(i2π/3) + e^(i4π/3) = cos(2π/3) + i sin(2π/3) + cos(4π/3) + i sin(4π/3) = (-1/2 + i√3/2) + (-1/2 - i√3/2) = -1. Therefore, adding 1 gives -1 + 1 = 0. So yes, that works. Similarly for the other roots. Therefore, all four roots are 6th roots of unity. Therefore, n=6? But the problem is asking for the smallest n such that all roots are nth roots of unity. Is 6 the minimal such n?

Wait, but let me check if maybe they are roots of unity of smaller order. For example, is e^(iπ/3) a 3rd root of unity? Let me see. The 3rd roots of unity are e^(i0), e^(i2π/3), e^(i4π/3). So e^(iπ/3) is not one of them. Similarly, e^(iπ/3) is a primitive 6th root of unity. Let me check the order. The order of e^(iπ/3) is the smallest positive integer k such that (e^(iπ/3))^k = 1. That would require k*(π/3) = 2π*m for some integer m, so k/3 = 2m, so k = 6m. Therefore, the minimal k is 6. Therefore, e^(iπ/3) is a primitive 6th root of unity. Similarly for the others. For example, e^(i2π/3) is a primitive 3rd root of unity? Wait, let me check. The order of e^(i2π/3) is the minimal k such that (e^(i2π/3))^k = 1. That requires k*(2π/3) = 2π*m => 2k/3 = 2m => k/3 = m. Therefore, minimal k is 3. Therefore, e^(i2π/3) is a primitive 3rd root of unity. Similarly, e^(i4π/3) is also a primitive 3rd root of unity, and e^(i5π/3) is a primitive 6th root of unity. Therefore, the roots have orders 3 and 6. Therefore, the minimal n such that all roots are nth roots of unity must be the least common multiple of 3 and 6, which is 6. Therefore, n=6.

But wait, hold on. Let me check if maybe the roots are also roots of unity of a smaller order? For example, if there is an n smaller than 6 that works? Let me check n=3. If n=3, then the 3rd roots of unity are 1, e^(i2π/3), e^(i4π/3). But the roots of our equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). However, e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity. So n=3 is too small. Similarly, n=4? Let's see. The 4th roots of unity are 1, i, -1, -i. Our roots are at angles π/3, 2π/3, 4π/3, 5π/3. None of these angles are multiples of π/2 (since π/3 ≈ 60°, which is not 90°, 180°, etc.), so n=4 is too small. How about n=5? The 5th roots of unity have angles 2π/5, 4π/5, 6π/5, 8π/5, which are not equal to π/3, 2π/3, etc. So n=5 is out. n=6 seems to be the first candidate.

But maybe there's a smaller n than 6? Let me think again. Let me check n=12? No, 6 is smaller. Wait, perhaps there is a smaller n? Let me check n=6. But is there a smaller n? For example, n=6 is the LCM of 3 and 6. But maybe if all roots have orders dividing n. Since some roots have order 3 and some have order 6, then n must be a multiple of 6. Therefore, the minimal n is 6.

Wait, but let me check if maybe the roots can be expressed as roots of unity of order less than 6? For example, is there an n such that all these roots are nth roots of unity? Let me think about the minimal polynomial. The equation z⁴ + z² + 1 = 0. Let me see if this polynomial is related to cyclotomic polynomials. Let me recall that cyclotomic polynomials are minimal polynomials over integers for primitive roots of unity.

For example, Φ₃(z) = z² + z + 1, Φ₆(z) = z² - z + 1, Φ₁₂(z) = z⁴ - z² + 1, etc. Wait, let me check. The cyclotomic polynomial Φ_n(z) is the product of (z - ζ) where ζ ranges over the primitive nth roots of unity. So, perhaps our polynomial z⁴ + z² + 1 can be related to Φ₁₂(z) or something else?

Wait, let me note that z⁶ - 1 = (z³ - 1)(z³ + 1) = (z - 1)(z² + z + 1)(z + 1)(z² - z + 1). Therefore, Φ₃(z) is z² + z + 1, Φ₆(z) is z² - z + 1. Also, Φ₁₂(z) is the minimal polynomial for primitive 12th roots of unity. Let me compute Φ₁₂(z). The formula for cyclotomic polynomials is Φ_n(z) = ∏_{k=1, gcd(k,n)=1}^n (z - e^(2πik/n)). For n=12, the primitive roots are those k where gcd(k,12)=1, so k=1,5,7,11. Therefore, Φ₁₂(z) = (z - e^(iπ/6))(z - e^(i5π/6))(z - e^(i7π/6))(z - e^(i11π/6)).

Let me compute Φ₁₂(z). Alternatively, there is a formula: Φ_n(z) = ∏_{d|n} (z^{n/d} - 1)^{μ(d)} where μ is the Möbius function. But maybe it's easier to note that Φ₁₂(z) divides z^φ(12) - 1, but φ(12)=4. Wait, no, Φ_n(z) divides z^n - 1, and the product of Φ_d(z) over d|n is z^n - 1. So, for n=12, z^12 - 1 = Φ₁(z)Φ₂(z)Φ₃(z)Φ₄(z)Φ₆(z)Φ₁₂(z). Since Φ₁(z)=z-1, Φ₂(z)=z+1, Φ₃(z)=z² + z +1, Φ₄(z)=z² + 1, Φ₆(z)=z² - z +1, and Φ₁₂(z) is the remaining factor.

So, Φ₁₂(z) = (z^12 -1)/[(z -1)(z +1)(z² + z +1)(z² +1)(z² - z +1)]. Let me compute that.

First, compute (z -1)(z +1)= z² -1.

Then multiply by (z² + z +1): (z² -1)(z² + z +1) = z^4 + z³ + z² - z² - z -1 = z^4 + z³ - z -1.

Then multiply by (z² +1): (z^4 + z³ - z -1)(z² +1) = z^6 + z^5 - z³ - z + z^4 + z³ - z -1 = z^6 + z^5 + z^4 - 2z -1. Wait, maybe I need to do this more carefully.

Alternatively, perhaps there's another approach. Let me recall that Φ₁₂(z) is z^4 - z² + 1. Wait, let me check with some known values. If z^6 -1 = (z^2 -1)(z^4 + z^2 +1). Wait, no:

Wait z^6 -1 = (z^3)^2 - 1 = (z^3 -1)(z^3 +1) = (z -1)(z^2 + z +1)(z +1)(z^2 - z +1). So, z^4 + z^2 +1 is not directly a factor here. Wait, but if I take z^6 -1 divided by (z^2 -1), which is Φ₁(z)Φ₂(z), I get z^4 + z² + 1? Let me check:

(z^6 -1) / (z^2 -1) = z^4 + z² +1. Yes! Because (z^2 -1)(z^4 + z² +1) = z^6 + z^4 + z² - z^4 - z² -1 = z^6 -1. Therefore, z^4 + z² +1 = (z^6 -1)/(z² -1). Therefore, z^4 + z² +1 is equal to Φ₃(z)Φ₆(z)? Wait, because z^6 -1 = Φ₁(z)Φ₂(z)Φ₃(z)Φ₆(z)Φ₁₂(z)? Wait, earlier we had z^6 -1 = (z^3 -1)(z^3 +1) = (z -1)(z² + z +1)(z +1)(z² - z +1). Which is Φ₁(z)Φ₃(z)Φ₂(z)Φ₆(z). Therefore, Φ₁₂(z) is not a factor here. Wait, but z^6 -1 is equal to z^4 + z² +1 multiplied by z² -1. Therefore, z^4 + z² +1 is equal to (z^6 -1)/(z² -1). Therefore, if we factor z^6 -1 as (z^2 -1)(z^4 + z² +1), then z^4 + z² +1 is equal to Φ₃(z)Φ₆(z)? Because z^6 -1 factors into Φ₁(z)Φ₂(z)Φ₃(z)Φ₆(z). So, Φ₁(z)Φ₂(z)Φ₃(z)Φ₆(z) = (z^2 -1)(z^4 + z² +1). So, Φ₁(z)Φ₂(z) = z² -1, and Φ₃(z)Φ₆(z) = z^4 + z² +1. Therefore, the polynomial z^4 + z² +1 factors into Φ₃(z)Φ₆(z). Therefore, the roots of z^4 + z² +1 =0 are the roots of Φ₃(z) and Φ₆(z). But Φ₃(z) has roots the primitive 3rd roots of unity, which are e^(i2π/3) and e^(i4π/3), and Φ₆(z) has roots the primitive 6th roots of unity, which are e^(iπ/3), e^(i5π/3), e^(iπ/3), e^(i5π/3)? Wait, no, Φ₆(z) is degree φ(6)=2. The primitive 6th roots of unity are e^(iπ/3) and e^(i5π/3). Therefore, Φ₆(z)=z² - z +1, which has roots e^(iπ/3) and e^(i5π/3). Wait, let me check. If Φ₆(z) is z² - z +1, then its roots satisfy z² - z +1 =0. Solving this quadratic equation: z = [1 ± √(-3)] / 2 = [1 ± i√3]/2. Which are e^(iπ/3) and e^(i5π/3), since e^(iπ/3) = cos(π/3) + i sin(π/3) = 1/2 + i√3/2 and e^(i5π/3) is 1/2 - i√3/2. Therefore, indeed Φ₆(z) = z² - z +1. Wait, but our original equation is z⁴ + z² +1 =0. So if it factors into Φ₃(z)Φ₆(z), then Φ₃(z) is z² + z +1 and Φ₆(z) is z² - z +1. Multiplying them gives (z² + z +1)(z² - z +1) = z^4 - z² + z² + z² - z + z + ... Wait, let me compute it properly. (z² + z +1)(z² - z +1) = z²(z² - z +1) + z(z² - z +1) +1(z² - z +1) = z^4 - z³ + z² + z³ - z² + z + z² - z +1. Then simplifying:

z^4 + (-z³ + z³) + (z² - z² + z²) + (z - z) + 1 = z^4 + z² + 1. Yes! Therefore, z⁴ + z² +1 factors as Φ₃(z)Φ₆(z). Therefore, the roots of z⁴ + z² +1 = 0 are precisely the primitive 3rd roots of unity and the primitive 6th roots of unity.

Therefore, the roots are the primitive 3rd roots of unity (which are order 3) and the primitive 6th roots of unity (which are order 6). Therefore, to include all these roots as nth roots of unity, n needs to be a multiple of both 3 and 6, so the least common multiple of 3 and 6 is 6. Therefore, the minimal n is 6. Therefore, the answer is 6?

But let me verify once more. For example, is 6 the minimal n? Suppose someone thinks maybe n=12? But 6 is smaller. Let me check if all roots are 6th roots of unity. The primitive 3rd roots of unity are also 6th roots of unity, since 3 divides 6, so they are 6th roots. Similarly, the primitive 6th roots of unity are 6th roots. Therefore, all roots are 6th roots of unity. Therefore, n=6 works. Since there is no smaller n (as we saw for n=3, 4, 5 do not work) then the minimal n is 6.

But let me just make sure there isn't a mistake here. For example, maybe the roots are also roots of unity of a smaller order. Wait, for example, if we have a root that is a primitive 6th root, then the order is 6, so n must be multiple of 6. The primitive 3rd roots have order 3, so n must be multiple of 3. Since 6 is the LCM of 3 and 6, that's the minimal n. Therefore, yes, n=6. Therefore, I think the answer is 6.

But just to be thorough, let me check if there exists a smaller n. Let's suppose n=6 is the minimal. Let me check n=4. As before, the 4th roots of unity are 1, -1, i, -i. But our roots are at angles π/3, 2π/3, 4π/3, 5π/3. None of these angles correspond to multiples of π/2 (which would be the angles for 4th roots), so no. Similarly, n=5: 5th roots of unity have angles 2π/5, 4π/5, 6π/5, 8π/5, 10π/5=2π. None of these are π/3 or 2π/3, etc. So n=5 is out. n=6 is the first where angles π/3, 2π/3, etc., are multiples of 2π/6=π/3. Wait, 2π/6=π/3. So yes, all the roots are 6th roots of unity. So n=6.

Alternatively, maybe there is an n less than 6? For example, n=12? But 6 is smaller. Wait, the question is about the minimal n. So 6 seems to be the answer.

Wait, but let me think again. Suppose that the roots are 6th roots of unity. But maybe they are 12th roots? But since 6 is smaller, the minimal n is 6. But let me check if maybe n=12 is required? For instance, if the roots have higher order? But no, we already saw that they are order 3 and 6, so n=6 suffices.

Alternatively, maybe I made a mistake in factoring the polynomial. Let me confirm that z⁴ + z² +1 factors into Φ₃(z)Φ₆(z). As we saw, Φ₃(z) is z² + z +1 and Φ₆(z) is z² - z +1. Multiplying them gives z⁴ + z² +1, which matches. Therefore, yes, the roots are exactly the primitive 3rd and 6th roots of unity. Therefore, n must be 6.

Alternatively, maybe the question is not asking for all roots to be primitive roots of unity but just roots of unity. But even if they are not primitive, as long as they are roots of unity, then they are included in some nth roots. But since the primitive ones have orders 3 and 6, then the nth must be multiple of 6, as we said.

Therefore, I think the answer is 6. But just to make sure, let me check with specific examples. Take the root e^(iπ/3), which is a primitive 6th root of unity. So it is a 6th root of unity. The other roots: e^(i2π/3) is a primitive 3rd root of unity, so it is also a 6th root of unity. Similarly for e^(i4π/3), e^(i5π/3). Therefore, all roots are 6th roots of unity. Therefore, n=6 works. Since there is no smaller n (as n=3 fails because of e^(iπ/3), n=4 fails as angles don't match, etc.), then n=6 is indeed minimal. Therefore, the answer is 6.

**Final Answer**
The smallest positive integer $ n $ is \boxed{6}.

To solve the problem, we are given the polynomial equation:

$$
z^4 + z^2 + 1 = 0
$$

We are asked to find the **smallest positive integer $ n $** such that **all the roots** of this equation are **$ n $th roots of unity**.

---

### Step 1: Understand the nature of the roots

We begin by recognizing that a **root of unity** is a complex number $ \omega $ satisfying:

$$
\omega^n = 1
$$

So, we first identify all the roots of the given polynomial. Let's denote the roots as $ z_1, z_2, z_3, z_4 $.

Let's rewrite the equation:

$$
z^4 + z^2 + 1 = 0
$$

We make a substitution: Let $ w = z^2 $, so that the equation becomes:

$$
w^2 + w + 1 = 0
$$

This is a quadratic in $ w $, with complex solutions:

$$
w = \frac{-1 \pm i\sqrt{3}}{2}
$$

Observe that these are the **primitive 3rd roots of unity**: $ e^{2\pi i / 3} $ and $ e^{4\pi i / 3} $.

Now, since $ w = z^2 $, the roots $ z $ of the original equation are the **square roots** of the above $ w $ values. Specifically:

- $ z = \sqrt{e^{2\pi i / 3}} = e^{\pi i / 3}, e^{4\pi i / 3} $
- $ z = \sqrt{e^{4\pi i / 3}} = e^{2\pi i / 3}, e^{5\pi i / 3} $

Thus, the four roots of $ z^4 + z^2 + 1 = 0 $ are:

$$
z = e^{i\pi/3},\ e^{i2\pi/3},\ e^{i4\pi/3},\ e^{i5\pi/3}
$$

---

### Step 2: Determine which roots are roots of unity

All these roots lie on the unit circle, and their angles are rational multiples of $ \pi $. Thus, they are **roots of unity**.

But to be precise, we consider their **order** — that is, the smallest positive integer $ n $ for which $ z^n = 1 $.

Let's check each root:

- $ e^{i\pi/3} $: this is a **primitive 6th root of unity**, since $ \pi/3 = 2\pi/6 $
- $ e^{i2\pi/3} $: a **primitive 3rd root of unity**, since $ 2\pi/3 = 2\pi/3 $
- $ e^{i4\pi/3} $: same as above
- $ e^{i5\pi/3} $: same as $ e^{i\pi/3} $, also a **primitive 6th root of unity**

Therefore, the roots have **orders 3 and 6**. This means all roots are **6th roots of unity**, because the 3rd roots of unity are also 6th roots of unity (since 3 divides 6).

---

### Step 3: Find the minimal such $ n $

The problem asks for the **smallest** $ n $ such that **all** roots are $ n $th roots of unity.

We now consider the orders of the roots:
- One root has **order 3**
- Others have **order 6**

So, $ n $ must be a **multiple of both 3 and 6**, which means $ n $ must be a **multiple of the least common multiple (LCM)** of these numbers.

$$
\text{LCM}(3, 6) = 6
$$

We verify that $ n = 6 $ works:
- All roots are 6th roots of unity (since both 3 and 6 divide 6)
- $ n = 6 $ is smaller than the next candidate (e.g., $ n = 12 $, or $ n = 3 $, which does not include all roots), so it’s **minimal**

---

### Final Conclusion

All the roots of the polynomial $ z^4 + z^2 + 1 = 0 $ are 6th roots of unity, and 6 is the **smallest** positive integer for which this is true.

---

### Final Answer

$$
\boxed{6}
$$

Solution:
Multiplying the equation $z^4 + z^2 + 1 = 0$ by $z^2 - 1 = (z - 1)(z + 1)$, we get $z^6 - 1 = 0$.  Therefore, every root of $z^4 + z^2 + 1 = 0$ is a sixth root of unity.

The sixth roots of unity are $e^{0}$, $e^{2 \pi i/6}$, $e^{4 \pi i/6}$, $e^{6 \pi i/6}$, $e^{8 \pi i/6}$, and $e^{10 \pi i/6}$.  We see that $e^{0} = 1$ and $e^{6 \pi i/6} = e^{\pi i} = -1$, so the roots of
\[z^4 + z^2 + 1 = 0\]are the remaining sixth roots of unity, namely $e^{2 \pi i/6}$, $e^{4 \pi i/6}$, $e^{8 \pi i/6}$, and $e^{10 \pi i/6}$.  The complex number $e^{2 \pi i/6}$ is a primitive sixth root of unity, so by definition, the smallest positive integer $n$ such that $(e^{2 \pi i/6})^n = 1$ is 6.  Therefore, the smallest possible value of $n$ is $\boxed{6}$.


Explanation:
Okay, so I need to figure out what power of 4 equals 8. Hmm, let me think. The question is asking for an exponent such that when 4 is raised to that exponent, the result is 8. In other words, I need to solve the equation 4^x = 8 for x.

First, I remember that exponents can sometimes be tricky, but maybe I can express both numbers with the same base? Let me recall that 4 and 8 are both powers of 2. Let me check that. 4 is 2 squared, right? So 4 = 2^2. And 8 is 2 cubed, so 8 = 2^3. Oh, that seems useful. If I can rewrite both sides of the equation with base 2, maybe I can solve for x more easily.

So let me rewrite the original equation 4^x = 8 using base 2. Since 4 is 2 squared, then 4^x is (2^2)^x. And then 8 is 2^3. So substituting these in, the equation becomes (2^2)^x = 2^3.

Now, I remember that when you raise a power to another power, you multiply the exponents. So (2^2)^x is the same as 2^(2*x). So the equation simplifies to 2^(2x) = 2^3.

Since the bases are the same (both are base 2), the exponents must be equal for the expressions to be equal. Therefore, 2x must equal 3.

So, solving for x, I divide both sides of the equation by 2. That gives me x = 3/2.

Wait, so is the answer 3/2? Let me check my work to make sure I didn't make a mistake.

If x is 3/2, then 4^(3/2) should equal 8. Let me calculate that.

First, 4^(1/2) is the square root of 4, which is 2. Then, 4^(3/2) is the same as (4^(1/2))^3, which is 2^3. And 2^3 is 8. Yep, that checks out.

Alternatively, I could think of 4^(3/2) as (4^3)^(1/2). Let's see, 4^3 is 64, and the square root of 64 is 8. That also works. So both methods confirm that 4^(3/2) is 8.

Therefore, the power of 4 that equals 8 is 3/2.

Wait, but the question says "Express your answer as a common fraction." Well, 3/2 is already a common fraction, so that should be the answer.

Alternatively, maybe there's another way to approach this problem? Let me try using logarithms. If I take the logarithm of both sides, maybe I can solve for x.

Starting with 4^x = 8. Taking the natural logarithm of both sides: ln(4^x) = ln(8). Using the logarithm power rule, that becomes x*ln(4) = ln(8). Then, solving for x gives x = ln(8)/ln(4).

Let me compute that. I know that ln(8) is ln(2^3) = 3*ln(2) and ln(4) is ln(2^2) = 2*ln(2). Therefore, x = (3*ln(2))/(2*ln(2)) = 3/2. So again, I get x = 3/2.

Same result. So that confirms it again.

Alternatively, if I use common logarithms, base 10, the same thing would happen. log(4^x) = log(8), so x*log(4) = log(8), x = log(8)/log(4). Which again, since 8 is 2^3 and 4 is 2^2, log(8) is 3*log(2) and log(4) is 2*log(2), so again x is 3/2.

So regardless of the method, I end up with 3/2.

Let me think if there's another way someone might approach this problem and maybe get confused. For example, someone might think, "Well, 4 squared is 16, which is bigger than 8, so maybe the exponent is between 1 and 2?" Which is true, and 3/2 is 1.5, so that's between 1 and 2. So that makes sense.

Alternatively, someone might think of 4^x = 8 as 4^x = 4^(something). Since 8 is not a power of 4, but as I showed earlier, expressing both numbers as powers of 2 allows us to solve the equation.

Another possible mistake: maybe someone might think that since 4 is 2 squared, then to get 8, which is 2 cubed, you need to add exponents or something. But that's not the case. The exponent rules are multiplicative when dealing with powers.

Wait, so if I think about 4^x = 8, and knowing that 4 is 2 squared, so substituting that in, (2^2)^x = 2^3. Then, using the power of a power rule, multiply the exponents: 2^(2x) = 2^3. Therefore, 2x = 3, so x = 3/2. Yep, same as before.

So I think that's solid. I don't see any mistakes here.

Alternatively, if someone isn't comfortable with exponents, they might try to write out the powers of 4 and see when they reach 8. Let's try that.

4^1 = 4

4^2 = 16

So between 1 and 2, 4^x is 8. But 4^1 is 4, 4^2 is 16. So 8 is in between. So to find the exponent, we need to find a fractional exponent between 1 and 2.

Alternatively, maybe use a logarithm table? But that's probably more complicated.

Alternatively, think of it as a square root. Since 4^(1/2) is 2, which is 2^1. Then 4^(3/2) is 4^(1 + 1/2) = 4^1 * 4^(1/2) = 4 * 2 = 8. So that works.

Therefore, 4^(3/2) is 8, so the answer is 3/2.

I think that's thorough enough. I can't think of any other way to approach this. All methods point to the answer being 3/2, which is a common fraction.

Just to recap:

Original problem: Find x such that 4^x = 8.

Express both sides as powers of 2:

4 = 2^2, 8 = 2^3.

So (2^2)^x = 2^3 => 2^(2x) = 2^3 => 2x = 3 => x = 3/2.

Alternatively, using logarithms, same result. Testing the value, 4^(3/2) is indeed 8.

Therefore, the answer is 3/2.

**Final Answer**
The power of 4 that equals 8 is \boxed{\dfrac{3}{2}}.

To solve the problem of finding what power of 4 is equal to 8, we need to determine the value of $ x $ such that:

$$
4^x = 8
$$

---

### Step 1: Express Both Numbers with the Same Base

We know that both 4 and 8 are powers of 2:

$$
4 = 2^2 \quad \text{and} \quad 8 = 2^3
$$

So we can rewrite the original equation in terms of base 2:

$$
(2^2)^x = 2^3
$$

---

### Step 2: Apply Exponent Rules

Using the **power of a power** rule, $(a^m)^n = a^{m \cdot n}$, we simplify the left-hand side:

$$
2^{2x} = 2^3
$$

Since the bases are the same, we can equate the exponents:

$$
2x = 3
$$

Solving for $ x $:

$$
x = \frac{3}{2}
$$

---

### Step 3: Verify the Solution

Let’s check the result:

$$
4^{3/2} = (2^2)^{3/2} = 2^{2 \cdot (3/2)} = 2^3 = 8
$$

This confirms our solution is correct.

---

### Final Answer

The power of 4 that equals 8 is:

$$
\boxed{\dfrac{3}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what power of 4 equals 8. Hmm, let me think. The question is asking for an exponent such that when 4 is raised to that exponent, the result is 8. In other words, I need to solve the equation 4^x = 8 for x.

First, I remember that exponents can sometimes be tricky, but maybe I can express both numbers with the same base? Let me recall that 4 and 8 are both powers of 2. Let me check that. 4 is 2 squared, right? So 4 = 2^2. And 8 is 2 cubed, so 8 = 2^3. Oh, that seems useful. If I can rewrite both sides of the equation with base 2, maybe I can solve for x more easily.

So let me rewrite the original equation 4^x = 8 using base 2. Since 4 is 2 squared, then 4^x is (2^2)^x. And then 8 is 2^3. So substituting these in, the equation becomes (2^2)^x = 2^3.

Now, I remember that when you raise a power to another power, you multiply the exponents. So (2^2)^x is the same as 2^(2*x). So the equation simplifies to 2^(2x) = 2^3.

Since the bases are the same (both are base 2), the exponents must be equal for the expressions to be equal. Therefore, 2x must equal 3.

So, solving for x, I divide both sides of the equation by 2. That gives me x = 3/2.

Wait, so is the answer 3/2? Let me check my work to make sure I didn't make a mistake.

If x is 3/2, then 4^(3/2) should equal 8. Let me calculate that.

First, 4^(1/2) is the square root of 4, which is 2. Then, 4^(3/2) is the same as (4^(1/2))^3, which is 2^3. And 2^3 is 8. Yep, that checks out.

Alternatively, I could think of 4^(3/2) as (4^3)^(1/2). Let's see, 4^3 is 64, and the square root of 64 is 8. That also works. So both methods confirm that 4^(3/2) is 8.

Therefore, the power of 4 that equals 8 is 3/2.

Wait, but the question says "Express your answer as a common fraction." Well, 3/2 is already a common fraction, so that should be the answer.

Alternatively, maybe there's another way to approach this problem? Let me try using logarithms. If I take the logarithm of both sides, maybe I can solve for x.

Starting with 4^x = 8. Taking the natural logarithm of both sides: ln(4^x) = ln(8). Using the logarithm power rule, that becomes x*ln(4) = ln(8). Then, solving for x gives x = ln(8)/ln(4).

Let me compute that. I know that ln(8) is ln(2^3) = 3*ln(2) and ln(4) is ln(2^2) = 2*ln(2). Therefore, x = (3*ln(2))/(2*ln(2)) = 3/2. So again, I get x = 3/2.

Same result. So that confirms it again.

Alternatively, if I use common logarithms, base 10, the same thing would happen. log(4^x) = log(8), so x*log(4) = log(8), x = log(8)/log(4). Which again, since 8 is 2^3 and 4 is 2^2, log(8) is 3*log(2) and log(4) is 2*log(2), so again x is 3/2.

So regardless of the method, I end up with 3/2.

Let me think if there's another way someone might approach this problem and maybe get confused. For example, someone might think, "Well, 4 squared is 16, which is bigger than 8, so maybe the exponent is between 1 and 2?" Which is true, and 3/2 is 1.5, so that's between 1 and 2. So that makes sense.

Alternatively, someone might think of 4^x = 8 as 4^x = 4^(something). Since 8 is not a power of 4, but as I showed earlier, expressing both numbers as powers of 2 allows us to solve the equation.

Another possible mistake: maybe someone might think that since 4 is 2 squared, then to get 8, which is 2 cubed, you need to add exponents or something. But that's not the case. The exponent rules are multiplicative when dealing with powers.

Wait, so if I think about 4^x = 8, and knowing that 4 is 2 squared, so substituting that in, (2^2)^x = 2^3. Then, using the power of a power rule, multiply the exponents: 2^(2x) = 2^3. Therefore, 2x = 3, so x = 3/2. Yep, same as before.

So I think that's solid. I don't see any mistakes here.

Alternatively, if someone isn't comfortable with exponents, they might try to write out the powers of 4 and see when they reach 8. Let's try that.

4^1 = 4

4^2 = 16

So between 1 and 2, 4^x is 8. But 4^1 is 4, 4^2 is 16. So 8 is in between. So to find the exponent, we need to find a fractional exponent between 1 and 2.

Alternatively, maybe use a logarithm table? But that's probably more complicated.

Alternatively, think of it as a square root. Since 4^(1/2) is 2, which is 2^1. Then 4^(3/2) is 4^(1 + 1/2) = 4^1 * 4^(1/2) = 4 * 2 = 8. So that works.

Therefore, 4^(3/2) is 8, so the answer is 3/2.

I think that's thorough enough. I can't think of any other way to approach this. All methods point to the answer being 3/2, which is a common fraction.

Just to recap:

Original problem: Find x such that 4^x = 8.

Express both sides as powers of 2:

4 = 2^2, 8 = 2^3.

So (2^2)^x = 2^3 => 2^(2x) = 2^3 => 2x = 3 => x = 3/2.

Alternatively, using logarithms, same result. Testing the value, 4^(3/2) is indeed 8.

Therefore, the answer is 3/2.

**Final Answer**
The power of 4 that equals 8 is \boxed{\dfrac{3}{2}}.

To solve the problem of finding what power of 4 is equal to 8, we need to determine the value of $ x $ such that:

$$
4^x = 8
$$

---

### Step 1: Express Both Numbers with the Same Base

We know that both 4 and 8 are powers of 2:

$$
4 = 2^2 \quad \text{and} \quad 8 = 2^3
$$

So we can rewrite the original equation in terms of base 2:

$$
(2^2)^x = 2^3
$$

---

### Step 2: Apply Exponent Rules

Using the **power of a power** rule, $(a^m)^n = a^{m \cdot n}$, we simplify the left-hand side:

$$
2^{2x} = 2^3
$$

Since the bases are the same, we can equate the exponents:

$$
2x = 3
$$

Solving for $ x $:

$$
x = \frac{3}{2}
$$

---

### Step 3: Verify the Solution

Let’s check the result:

$$
4^{3/2} = (2^2)^{3/2} = 2^{2 \cdot (3/2)} = 2^3 = 8
$$

This confirms our solution is correct.

---

### Final Answer

The power of 4 that equals 8 is:

$$
\boxed{\dfrac{3}{2}}
$$

Solution: We are asked to solve $4^x=8$ for $x$.  Writing $4$ as $2^2$ and $8$ as $2^3$, the equation becomes $(2^2)^x=2^3$.  The left-hand side simplifies to $2^{2x}$, so we may set exponents equal to find $2x=3$, which implies $x=\boxed{\frac{3}{2}}$.


Explanation:
Okay, so I need to find the quotient when x⁶ - 3 is divided by x + 1. Hmm, let me think. I remember that dividing polynomials can be done using long division, or maybe synthetic division? I think synthetic division is quicker, but I need to make sure it applies here. Wait, synthetic division works when dividing by a linear factor, right? And x + 1 is a linear factor, so that should work. Let me recall how synthetic division works.

First, for synthetic division, you need to use the root of the divisor. The divisor is x + 1, so the root is -1. Then, I need to write down the coefficients of the polynomial x⁶ - 3. Let me check: the polynomial is x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3. So the coefficients are 1, 0, 0, 0, 0, 0, -3.

Okay, so setting up synthetic division with -1. Let me write that down:

-1 | 1   0   0   0   0   0   -3

Now, I need to bring down the first coefficient, which is 1. Then multiply it by -1, which gives -1. Add that to the next coefficient (which is 0), resulting in -1. Then multiply that -1 by -1 to get 1, add to the next coefficient (0), so 1. Multiply by -1 again to get -1, add to next 0: -1. Multiply by -1 to get 1, add to next 0: 1. Multiply by -1 to get -1, add to next 0: -1. Multiply by -1 to get 1, add to the last coefficient, -3: 1 + (-3) = -2.

So the numbers I have are: 1, -1, 1, -1, 1, -1, -2. Wait, so the quotient would be the polynomial of degree 5 with coefficients 1, -1, 1, -1, 1, -1, and the remainder is -2. Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x - 1, and the remainder is -2.

Wait, let me confirm this. Because if I multiply (x + 1) by the quotient, I should get the original polynomial plus the remainder. Let me check that.

So, (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + (-2). Let me expand the product:

First, multiply x by each term of the quotient:

x*(x⁵) = x⁶

x*(-x⁴) = -x⁵

x*(x³) = x⁴

x*(-x²) = -x³

x*(x) = x²

x*(-1) = -x

Then multiply 1 by each term of the quotient:

1*(x⁵) = x⁵

1*(-x⁴) = -x⁴

1*(x³) = x³

1*(-x²) = -x²

1*(x) = x

1*(-1) = -1

Now, add all these together:

x⁶ - x⁵ + x⁴ - x³ + x² - x + x⁵ - x⁴ + x³ - x² + x - 1

Now combine like terms:

x⁶ + (-x⁵ + x⁵) + (x⁴ - x⁴) + (-x³ + x³) + (x² - x²) + (-x + x) -1

Simplify each term:

x⁶ + 0 + 0 + 0 + 0 + 0 -1 = x⁶ - 1

Then adding the remainder: x⁶ - 1 + (-2) = x⁶ - 3. Which is exactly the original polynomial. So that checks out! Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1, and the remainder is -2. Since the question asks for the quotient, I just need to write that polynomial.

Alternatively, maybe I can verify using polynomial long division as well? Let me try that to confirm.

Dividing x⁶ - 3 by x + 1. Let's set it up:

Divide x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3 by x + 1.

First term: x⁶ divided by x is x⁵. Multiply x + 1 by x⁵: x⁶ + x⁵. Subtract that from the original polynomial:

(x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3) - (x⁶ + x⁵) = -x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3

Next term: -x⁵ divided by x is -x⁴. Multiply x + 1 by -x⁴: -x⁵ - x⁴. Subtract that:

(-x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3) - (-x⁵ - x⁴) = 0x⁵ + x⁴ + 0x³ + 0x² + 0x - 3

Next term: x⁴ / x = x³. Multiply x + 1 by x³: x⁴ + x³. Subtract:

(x⁴ + 0x³ + 0x² + 0x - 3) - (x⁴ + x³) = 0x⁴ - x³ + 0x² + 0x - 3

Next term: -x³ / x = -x². Multiply x + 1 by -x²: -x³ - x². Subtract:

(-x³ + 0x² + 0x - 3) - (-x³ - x²) = 0x³ + x² + 0x - 3

Next term: x² / x = x. Multiply x + 1 by x: x² + x. Subtract:

(x² + 0x - 3) - (x² + x) = 0x² - x - 3

Next term: -x / x = -1. Multiply x + 1 by -1: -x -1. Subtract:

(-x - 3) - (-x -1) = (-x -3) + x +1 = (-x + x) + (-3 +1) = 0 -2 = -2.

So the remainder is -2, and the quotient is x⁵ - x⁴ + x³ - x² + x -1. Which matches the result from synthetic division. Therefore, both methods confirm that the quotient is x⁵ - x⁴ + x³ - x² + x -1.

Alternatively, maybe there's another way? Like using the Remainder Theorem or something else? Wait, the Remainder Theorem says that the remainder when dividing by x - a is f(a). In this case, dividing by x + 1 is like dividing by x - (-1), so the remainder would be f(-1). Let's check that. f(-1) = (-1)^6 - 3 = 1 - 3 = -2. Which matches our previous results. So that gives me more confidence.

But since the question is asking for the quotient, not the remainder, so I need to focus on that. So the quotient is the polynomial we found, x⁵ - x⁴ + x³ - x² + x -1. Let me check if that can be simplified or factored further? Well, maybe there's a pattern here. The coefficients alternate between 1 and -1, except the last term is -1. Wait, let me write it again:

x⁵ - x⁴ + x³ - x² + x - 1. If I factor this, perhaps? Let me see:

It looks like it's a geometric series? Let me think. Let me recall that x⁶ - 1 can be factored as (x - 1)(x⁵ + x⁴ + x³ + x² + x + 1). But our polynomial is different. However, in our case, we have x⁶ - 3. But maybe that's not directly helpful. Alternatively, since we divided x⁶ - 3 by x + 1, maybe we can relate it to x⁶ - 1?

Wait, if I consider x⁶ - 3 = (x⁶ - 1) - 2. Then, since x⁶ - 1 factors as (x + 1)(x⁵ - x⁴ + x³ - x² + x -1) + 0? Wait, earlier we saw that (x + 1)(quotient) = x⁶ - 1. Wait, no, earlier when we did the multiplication, we had (x + 1)(quotient) = x⁶ - 1. Then, since our original polynomial is x⁶ - 3 = (x⁶ - 1) - 2, then dividing by x + 1 would give the same quotient as (x⁶ - 1)/(x + 1) minus 2/(x + 1). But since we are doing polynomial division, the remainder is -2. So this also shows that the quotient is the same as (x⁶ - 1)/(x + 1), which is x⁵ - x⁴ + x³ - x² + x -1. Therefore, this is consistent.

Therefore, all methods lead to the same quotient. Therefore, I can be confident that the quotient is x⁵ - x⁴ + x³ - x² + x - 1.

Wait, but let me check if the quotient is correct by plugging in a value for x and see if the original polynomial equals (x + 1)*quotient + remainder. Let me pick x = 1. Then original polynomial is 1⁶ - 3 = 1 - 3 = -2. The quotient evaluated at x = 1 is 1 - 1 + 1 -1 +1 -1 = 0. Then (1 + 1)*0 + (-2) = 0 -2 = -2, which matches. Let me try x = 0. Original polynomial: 0 - 3 = -3. Quotient at x=0 is 0 -0 +0 -0 +0 -1 = -1. Then (0 + 1)*(-1) + (-2) = -1 -2 = -3, which matches. Another test: x = 2. Original polynomial: 2⁶ - 3 = 64 - 3 = 61. Quotient at x=2: 32 - 16 + 8 - 4 + 2 -1 = Let's compute step by step: 32 -16 = 16; 16 +8=24; 24 -4=20; 20 +2=22; 22 -1=21. Then (2 + 1)*21 + (-2) = 3*21 -2 = 63 - 2 = 61. Which matches. So that checks out. Therefore, my quotient is correct.

Therefore, the answer is x⁵ - x⁴ + x³ - x² + x -1. So written in boxed form, it would be \boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.

**Final Answer**
The quotient is \boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.

To find the quotient when $ x^6 - 3 $ is divided by $ x + 1 $, we approach this using **synthetic division**, which is a streamlined method for dividing a polynomial by a binomial of the form $ x - a $, or in this case, $ x + 1 $, which is equivalent to $ x - (-1) $.

---

### Step 1: Set up the synthetic division

We begin by identifying the root of the divisor $ x + 1 $, which is $ -1 $. Then, we write down the coefficients of the polynomial $ x^6 - 3 $, which can be fully expanded as:

$$
x^6 - 3 = x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 3
$$

So, the coefficients are:
$$
1 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad -3
$$

---

### Step 2: Perform synthetic division

We perform the synthetic division with the root $ -1 $:

```
-1 |  1   0   0   0   0   0   -3
    |     -1   1  -1   1  -1   1
    ----------------------------
      1  -1   1  -1   1  -1   -2
```

The resulting row (excluding the last value, which is the **remainder**) gives us the coefficients of the **quotient**:

$$
1, -1, 1, -1, 1, -1
$$

So, the quotient polynomial is:

$$
x^5 - x^4 + x^3 - x^2 + x - 1
$$

The last number in the synthetic division setup, which is **-2**, is the **remainder**.

---

### Step 3: Verify the result

To ensure the quotient is correct, we can reconstruct the original polynomial using the quotient and the divisor:

$$
(x + 1)(x^5 - x^4 + x^3 - x^2 + x - 1) + (-2)
$$

Multiplying out the terms, we get:

$$
(x + 1)(x^5 - x^4 + x^3 - x^2 + x - 1) = x^6 - x^5 + x^4 - x^3 + x^2 - x + x^5 - x^4 + x^3 - x^2 + x - 1 = x^6 - 1
$$

Adding the remainder:

$$
x^6 - 1 + (-2) = x^6 - 3
$$

Which matches the original polynomial. So, the quotient is **correct**.

---

### Final Answer

$$
\boxed{x^5 - x^4 + x^3 - x^2 + x - 1}
$$

[Assistant]
Explanation:
Okay, so I need to find the quotient when x⁶ - 3 is divided by x + 1. Hmm, let me think. I remember that dividing polynomials can be done using long division, or maybe synthetic division? I think synthetic division is quicker, but I need to make sure it applies here. Wait, synthetic division works when dividing by a linear factor, right? And x + 1 is a linear factor, so that should work. Let me recall how synthetic division works.

First, for synthetic division, you need to use the root of the divisor. The divisor is x + 1, so the root is -1. Then, I need to write down the coefficients of the polynomial x⁶ - 3. Let me check: the polynomial is x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3. So the coefficients are 1, 0, 0, 0, 0, 0, -3.

Okay, so setting up synthetic division with -1. Let me write that down:

-1 | 1   0   0   0   0   0   -3

Now, I need to bring down the first coefficient, which is 1. Then multiply it by -1, which gives -1. Add that to the next coefficient (which is 0), resulting in -1. Then multiply that -1 by -1 to get 1, add to the next coefficient (0), so 1. Multiply by -1 again to get -1, add to next 0: -1. Multiply by -1 to get 1, add to next 0: 1. Multiply by -1 to get -1, add to next 0: -1. Multiply by -1 to get 1, add to the last coefficient, -3: 1 + (-3) = -2.

So the numbers I have are: 1, -1, 1, -1, 1, -1, -2. Wait, so the quotient would be the polynomial of degree 5 with coefficients 1, -1, 1, -1, 1, -1, and the remainder is -2. Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x - 1, and the remainder is -2.

Wait, let me confirm this. Because if I multiply (x + 1) by the quotient, I should get the original polynomial plus the remainder. Let me check that.

So, (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + (-2). Let me expand the product:

First, multiply x by each term of the quotient:

x*(x⁵) = x⁶

x*(-x⁴) = -x⁵

x*(x³) = x⁴

x*(-x²) = -x³

x*(x) = x²

x*(-1) = -x

Then multiply 1 by each term of the quotient:

1*(x⁵) = x⁵

1*(-x⁴) = -x⁴

1*(x³) = x³

1*(-x²) = -x²

1*(x) = x

1*(-1) = -1

Now, add all these together:

x⁶ - x⁵ + x⁴ - x³ + x² - x + x⁵ - x⁴ + x³ - x² + x - 1

Now combine like terms:

x⁶ + (-x⁵ + x⁵) + (x⁴ - x⁴) + (-x³ + x³) + (x² - x²) + (-x + x) -1

Simplify each term:

x⁶ + 0 + 0 + 0 + 0 + 0 -1 = x⁶ - 1

Then adding the remainder: x⁶ - 1 + (-2) = x⁶ - 3. Which is exactly the original polynomial. So that checks out! Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1, and the remainder is -2. Since the question asks for the quotient, I just need to write that polynomial.

Alternatively, maybe I can verify using polynomial long division as well? Let me try that to confirm.

Dividing x⁶ - 3 by x + 1. Let's set it up:

Divide x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3 by x + 1.

First term: x⁶ divided by x is x⁵. Multiply x + 1 by x⁵: x⁶ + x⁵. Subtract that from the original polynomial:

(x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3) - (x⁶ + x⁵) = -x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3

Next term: -x⁵ divided by x is -x⁴. Multiply x + 1 by -x⁴: -x⁵ - x⁴. Subtract that:

(-x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3) - (-x⁵ - x⁴) = 0x⁵ + x⁴ + 0x³ + 0x² + 0x - 3

Next term: x⁴ / x = x³. Multiply x + 1 by x³: x⁴ + x³. Subtract:

(x⁴ + 0x³ + 0x² + 0x - 3) - (x⁴ + x³) = 0x⁴ - x³ + 0x² + 0x - 3

Next term: -x³ / x = -x². Multiply x + 1 by -x²: -x³ - x². Subtract:

(-x³ + 0x² + 0x - 3) - (-x³ - x²) = 0x³ + x² + 0x - 3

Next term: x² / x = x. Multiply x + 1 by x: x² + x. Subtract:

(x² + 0x - 3) - (x² + x) = 0x² - x - 3

Next term: -x / x = -1. Multiply x + 1 by -1: -x -1. Subtract:

(-x - 3) - (-x -1) = (-x -3) + x +1 = (-x + x) + (-3 +1) = 0 -2 = -2.

So the remainder is -2, and the quotient is x⁵ - x⁴ + x³ - x² + x -1. Which matches the result from synthetic division. Therefore, both methods confirm that the quotient is x⁵ - x⁴ + x³ - x² + x -1.

Alternatively, maybe there's another way? Like using the Remainder Theorem or something else? Wait, the Remainder Theorem says that the remainder when dividing by x - a is f(a). In this case, dividing by x + 1 is like dividing by x - (-1), so the remainder would be f(-1). Let's check that. f(-1) = (-1)^6 - 3 = 1 - 3 = -2. Which matches our previous results. So that gives me more confidence.

But since the question is asking for the quotient, not the remainder, so I need to focus on that. So the quotient is the polynomial we found, x⁵ - x⁴ + x³ - x² + x -1. Let me check if that can be simplified or factored further? Well, maybe there's a pattern here. The coefficients alternate between 1 and -1, except the last term is -1. Wait, let me write it again:

x⁵ - x⁴ + x³ - x² + x - 1. If I factor this, perhaps? Let me see:

It looks like it's a geometric series? Let me think. Let me recall that x⁶ - 1 can be factored as (x - 1)(x⁵ + x⁴ + x³ + x² + x + 1). But our polynomial is different. However, in our case, we have x⁶ - 3. But maybe that's not directly helpful. Alternatively, since we divided x⁶ - 3 by x + 1, maybe we can relate it to x⁶ - 1?

Wait, if I consider x⁶ - 3 = (x⁶ - 1) - 2. Then, since x⁶ - 1 factors as (x + 1)(x⁵ - x⁴ + x³ - x² + x -1) + 0? Wait, earlier we saw that (x + 1)(quotient) = x⁶ - 1. Wait, no, earlier when we did the multiplication, we had (x + 1)(quotient) = x⁶ - 1. Then, since our original polynomial is x⁶ - 3 = (x⁶ - 1) - 2, then dividing by x + 1 would give the same quotient as (x⁶ - 1)/(x + 1) minus 2/(x + 1). But since we are doing polynomial division, the remainder is -2. So this also shows that the quotient is the same as (x⁶ - 1)/(x + 1), which is x⁵ - x⁴ + x³ - x² + x -1. Therefore, this is consistent.

Therefore, all methods lead to the same quotient. Therefore, I can be confident that the quotient is x⁵ - x⁴ + x³ - x² + x - 1.

Wait, but let me check if the quotient is correct by plugging in a value for x and see if the original polynomial equals (x + 1)*quotient + remainder. Let me pick x = 1. Then original polynomial is 1⁶ - 3 = 1 - 3 = -2. The quotient evaluated at x = 1 is 1 - 1 + 1 -1 +1 -1 = 0. Then (1 + 1)*0 + (-2) = 0 -2 = -2, which matches. Let me try x = 0. Original polynomial: 0 - 3 = -3. Quotient at x=0 is 0 -0 +0 -0 +0 -1 = -1. Then (0 + 1)*(-1) + (-2) = -1 -2 = -3, which matches. Another test: x = 2. Original polynomial: 2⁶ - 3 = 64 - 3 = 61. Quotient at x=2: 32 - 16 + 8 - 4 + 2 -1 = Let's compute step by step: 32 -16 = 16; 16 +8=24; 24 -4=20; 20 +2=22; 22 -1=21. Then (2 + 1)*21 + (-2) = 3*21 -2 = 63 - 2 = 61. Which matches. So that checks out. Therefore, my quotient is correct.

Therefore, the answer is x⁵ - x⁴ + x³ - x² + x -1. So written in boxed form, it would be \boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.

**Final Answer**
The quotient is \boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.

To find the quotient when $ x^6 - 3 $ is divided by $ x + 1 $, we approach this using **synthetic division**, which is a streamlined method for dividing a polynomial by a binomial of the form $ x - a $, or in this case, $ x + 1 $, which is equivalent to $ x - (-1) $.

---

### Step 1: Set up the synthetic division

We begin by identifying the root of the divisor $ x + 1 $, which is $ -1 $. Then, we write down the coefficients of the polynomial $ x^6 - 3 $, which can be fully expanded as:

$$
x^6 - 3 = x^6 + 0x^5 + 0x^4 + 0x^3 + 0x^2 + 0x - 3
$$

So, the coefficients are:
$$
1 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad -3
$$

---

### Step 2: Perform synthetic division

We perform the synthetic division with the root $ -1 $:

```
-1 |  1   0   0   0   0   0   -3
    |     -1   1  -1   1  -1   1
    ----------------------------
      1  -1   1  -1   1  -1   -2
```

The resulting row (excluding the last value, which is the **remainder**) gives us the coefficients of the **quotient**:

$$
1, -1, 1, -1, 1, -1
$$

So, the quotient polynomial is:

$$
x^5 - x^4 + x^3 - x^2 + x - 1
$$

The last number in the synthetic division setup, which is **-2**, is the **remainder**.

---

### Step 3: Verify the result

To ensure the quotient is correct, we can reconstruct the original polynomial using the quotient and the divisor:

$$
(x + 1)(x^5 - x^4 + x^3 - x^2 + x - 1) + (-2)
$$

Multiplying out the terms, we get:

$$
(x + 1)(x^5 - x^4 + x^3 - x^2 + x - 1) = x^6 - x^5 + x^4 - x^3 + x^2 - x + x^5 - x^4 + x^3 - x^2 + x - 1 = x^6 - 1
$$

Adding the remainder:

$$
x^6 - 1 + (-2) = x^6 - 3
$$

Which matches the original polynomial. So, the quotient is **correct**.

---

### Final Answer

$$
\boxed{x^5 - x^4 + x^3 - x^2 + x - 1}
$$

Solution:
We can perform long division.  Alternatively, by the Remainder Theorem, the remainder upon division is $(-1)^6 - 3 = -2.$  Thus, we can write
\begin{align*}
\frac{x^6 - 3}{x + 1} &= \frac{(x^6 - 1) - 2}{x + 1} \\
&= \frac{x^6 - 1}{x + 1} - \frac{2}{x + 1} \\
&= \frac{(x^3 - 1)(x^3 + 1)}{x + 1} - \frac{2}{x + 1} \\
&= \frac{(x^3 - 1)(x + 1)(x^2 - x + 1)}{x + 1} - \frac{2}{x + 1} \\
&= (x^3 - 1)(x^2 - x + 1) - \frac{2}{x + 1} \\
&= x^5 - x^4 + x^3 - x^2 + x - 1 - \frac{2}{x + 1}.
\end{align*}Thus, the quotient is $\boxed{x^5 - x^4 + x^3 - x^2 + x - 1}.$


Explanation:
Okay, so I need to compute arcsin(-1/2) and express the answer in radians. Hmm, let me recall what arcsin does. The arcsin function is the inverse of the sine function, right? But I have to remember that the domain of arcsin is between -1 and 1, and its range is restricted to [-π/2, π/2] radians. So, arcsin(x) gives me an angle in that interval whose sine is x.

So, the question is asking for an angle θ such that sin(θ) = -1/2, and θ is between -π/2 and π/2. Let me think. I know that sin(π/6) is 1/2. Wait, so if I have sin(θ) = -1/2, maybe θ is -π/6? Because sine is an odd function, so sin(-θ) = -sin(θ). So, if sin(π/6) = 1/2, then sin(-π/6) = -1/2.

But let me check if -π/6 is within the range of arcsin. The range of arcsin is from -π/2 to π/2, which is -90 degrees to 90 degrees. -π/6 is -30 degrees, which is within that range. So that seems okay.

But wait, is there another angle in the range [-π/2, π/2] where sine is -1/2? Let me visualize the unit circle. In the fourth quadrant, angles between -π/2 and 0 have negative sine values. So, the angle -π/6 is in the fourth quadrant. The reference angle here is π/6. But since sine is negative, the angle is in the fourth quadrant, which matches with -π/6.

Alternatively, if I think about positive angles, the angle π/6 is in the first quadrant, and then another angle in the third quadrant would be π + π/6 = 7π/6, but that's not in the range of arcsin because 7π/6 is greater than π/2. So, arcsin only gives me the angle in the range [-π/2, π/2], so the answer must be -π/6.

Let me confirm with another method. Let me recall some key values. The sine of π/6 is 1/2, as mentioned. So, sine of -π/6 would be -1/2. Therefore, arcsin(-1/2) is -π/6.

But wait, sometimes people might get confused between the principal value and other solutions. But since the question specifies to express the answer in radians and not mention anything else, I think the principal value is what they want here.

Another way to think about it: if I consider the unit circle, the sine of an angle is equal to the y-coordinate. So, looking for a y-coordinate of -1/2. The point on the unit circle where y = -1/2 would be at angles -π/6 and 7π/6, etc. But only -π/6 is in the range of arcsin.

Wait, 7π/6 is 210 degrees, which is in the third quadrant, but that's not in the range of arcsin. The arcsin function only returns angles between -π/2 and π/2. So, yes, -π/6 is the correct answer.

But let me check with a calculator to make sure. If I calculate arcsin(-1/2), does it give me -π/6? Well, if I convert π/6 to decimal, that's approximately 0.5236 radians. So, -π/6 would be approximately -0.5236 radians. Let me check using a calculator.

Assuming I don't have a calculator, but I know that sin(-π/6) is indeed -sin(π/6) = -1/2, so that's correct. Therefore, arcsin(-1/2) = -π/6.

Is there a chance that the answer might be expressed as a positive angle? For example, sometimes angles are expressed in terms of positive radians. But the range of arcsin is specifically from -π/2 to π/2, so negative angles are allowed. Therefore, -π/6 is the correct answer.

Alternatively, if the question had asked for all solutions, then there would be infinitely many, but since it's just asking for arcsin(-1/2), which is the principal value, it's definitely -π/6.

Wait, just to confirm, is -π/6 within the range? The range is from -π/2 (-1.5708...) to π/2 (1.5708...). -π/6 is approximately -0.5236, which is between -π/2 and π/2. Yes, so that's valid.

So, after all this thinking, I believe the answer is -π/6 radians. I don't see any mistakes in my reasoning. Let me just recap:

1. arcsin(x) gives an angle θ in [-π/2, π/2] such that sin(θ) = x.
2. We need θ where sin(θ) = -1/2.
3. The reference angle is π/6 because sin(π/6) = 1/2.
4. Since sine is negative, the angle is in the fourth quadrant, so θ = -π/6.
5. -π/6 is within the range of arcsin.
6. Therefore, the answer is -π/6.

Yep, that all checks out. I think that's solid.

**Final Answer**
The value of $\arcsin \left( -\frac{1}{2} \right)$ is \boxed{-\frac{\pi}{6}}.

To find the value of $\arcsin\left(-\frac{1}{2}\right)$, we begin by recalling the definition of the inverse sine function.

---

### Step 1: Understand the Function

The **arcsin** function, or inverse sine, is defined as follows:

$$
\arcsin(x) = \theta \iff \sin(\theta) = x \quad \text{and} \quad \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$

So, $\arcsin\left(-\frac{1}{2}\right)$ is the **angle in the interval** $[- \frac{\pi}{2}, \frac{\pi}{2}]$ such that the sine of that angle is $-\frac{1}{2}$.

---

### Step 2: Use Known Trigonometric Values

We know from basic trigonometry:

$$
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
$$

Since sine is an **odd function**, we have:

$$
\sin(-\theta) = -\sin(\theta) \Rightarrow \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}
$$

So,

$$
\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}
$$

---

### Step 3: Check Range

We must ensure that the angle $-\frac{\pi}{6}$ lies within the **domain of the arcsin function**, which is:

$$
\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$

Since $-\frac{\pi}{6} \approx -0.5236$ is within $[-1.5708, 1.5708]$, it is indeed a valid angle.

---

### Final Verification

We can also verify this by checking:

$$
\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}
$$

This confirms that the angle $-\frac{\pi}{6}$ satisfies the definition of the inverse sine function.

---

### Final Answer

$$
\boxed{-\frac{\pi}{6}}
$$

[Assistant]
Explanation:
Okay, so I need to compute arcsin(-1/2) and express the answer in radians. Hmm, let me recall what arcsin does. The arcsin function is the inverse of the sine function, right? But I have to remember that the domain of arcsin is between -1 and 1, and its range is restricted to [-π/2, π/2] radians. So, arcsin(x) gives me an angle in that interval whose sine is x.

So, the question is asking for an angle θ such that sin(θ) = -1/2, and θ is between -π/2 and π/2. Let me think. I know that sin(π/6) is 1/2. Wait, so if I have sin(θ) = -1/2, maybe θ is -π/6? Because sine is an odd function, so sin(-θ) = -sin(θ). So, if sin(π/6) = 1/2, then sin(-π/6) = -1/2.

But let me check if -π/6 is within the range of arcsin. The range of arcsin is from -π/2 to π/2, which is -90 degrees to 90 degrees. -π/6 is -30 degrees, which is within that range. So that seems okay.

But wait, is there another angle in the range [-π/2, π/2] where sine is -1/2? Let me visualize the unit circle. In the fourth quadrant, angles between -π/2 and 0 have negative sine values. So, the angle -π/6 is in the fourth quadrant. The reference angle here is π/6. But since sine is negative, the angle is in the fourth quadrant, which matches with -π/6.

Alternatively, if I think about positive angles, the angle π/6 is in the first quadrant, and then another angle in the third quadrant would be π + π/6 = 7π/6, but that's not in the range of arcsin because 7π/6 is greater than π/2. So, arcsin only gives me the angle in the range [-π/2, π/2], so the answer must be -π/6.

Let me confirm with another method. Let me recall some key values. The sine of π/6 is 1/2, as mentioned. So, sine of -π/6 would be -1/2. Therefore, arcsin(-1/2) is -π/6.

But wait, sometimes people might get confused between the principal value and other solutions. But since the question specifies to express the answer in radians and not mention anything else, I think the principal value is what they want here.

Another way to think about it: if I consider the unit circle, the sine of an angle is equal to the y-coordinate. So, looking for a y-coordinate of -1/2. The point on the unit circle where y = -1/2 would be at angles -π/6 and 7π/6, etc. But only -π/6 is in the range of arcsin.

Wait, 7π/6 is 210 degrees, which is in the third quadrant, but that's not in the range of arcsin. The arcsin function only returns angles between -π/2 and π/2. So, yes, -π/6 is the correct answer.

But let me check with a calculator to make sure. If I calculate arcsin(-1/2), does it give me -π/6? Well, if I convert π/6 to decimal, that's approximately 0.5236 radians. So, -π/6 would be approximately -0.5236 radians. Let me check using a calculator.

Assuming I don't have a calculator, but I know that sin(-π/6) is indeed -sin(π/6) = -1/2, so that's correct. Therefore, arcsin(-1/2) = -π/6.

Is there a chance that the answer might be expressed as a positive angle? For example, sometimes angles are expressed in terms of positive radians. But the range of arcsin is specifically from -π/2 to π/2, so negative angles are allowed. Therefore, -π/6 is the correct answer.

Alternatively, if the question had asked for all solutions, then there would be infinitely many, but since it's just asking for arcsin(-1/2), which is the principal value, it's definitely -π/6.

Wait, just to confirm, is -π/6 within the range? The range is from -π/2 (-1.5708...) to π/2 (1.5708...). -π/6 is approximately -0.5236, which is between -π/2 and π/2. Yes, so that's valid.

So, after all this thinking, I believe the answer is -π/6 radians. I don't see any mistakes in my reasoning. Let me just recap:

1. arcsin(x) gives an angle θ in [-π/2, π/2] such that sin(θ) = x.
2. We need θ where sin(θ) = -1/2.
3. The reference angle is π/6 because sin(π/6) = 1/2.
4. Since sine is negative, the angle is in the fourth quadrant, so θ = -π/6.
5. -π/6 is within the range of arcsin.
6. Therefore, the answer is -π/6.

Yep, that all checks out. I think that's solid.

**Final Answer**
The value of $\arcsin \left( -\frac{1}{2} \right)$ is \boxed{-\frac{\pi}{6}}.

To find the value of $\arcsin\left(-\frac{1}{2}\right)$, we begin by recalling the definition of the inverse sine function.

---

### Step 1: Understand the Function

The **arcsin** function, or inverse sine, is defined as follows:

$$
\arcsin(x) = \theta \iff \sin(\theta) = x \quad \text{and} \quad \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$

So, $\arcsin\left(-\frac{1}{2}\right)$ is the **angle in the interval** $[- \frac{\pi}{2}, \frac{\pi}{2}]$ such that the sine of that angle is $-\frac{1}{2}$.

---

### Step 2: Use Known Trigonometric Values

We know from basic trigonometry:

$$
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
$$

Since sine is an **odd function**, we have:

$$
\sin(-\theta) = -\sin(\theta) \Rightarrow \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}
$$

So,

$$
\arcsin\left(-\frac{1}{2}\right) = -\frac{\pi}{6}
$$

---

### Step 3: Check Range

We must ensure that the angle $-\frac{\pi}{6}$ lies within the **domain of the arcsin function**, which is:

$$
\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
$$

Since $-\frac{\pi}{6} \approx -0.5236$ is within $[-1.5708, 1.5708]$, it is indeed a valid angle.

---

### Final Verification

We can also verify this by checking:

$$
\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}
$$

This confirms that the angle $-\frac{\pi}{6}$ satisfies the definition of the inverse sine function.

---

### Final Answer

$$
\boxed{-\frac{\pi}{6}}
$$

Solution: Since $\sin \left( -\frac{\pi}{6} \right) = -\frac{1}{2},$ $\arcsin \left( -\frac{1}{2} \right) = \boxed{-\frac{\pi}{6}}.$


Explanation:
Okay, so I need to convert the fraction 3/20 into a decimal. Hmm, let me think about how to do that. I remember that converting fractions to decimals usually involves division. So maybe I need to divide 3 by 20? Let me try that.

Alright, let's set it up. If I divide 3 by 20, that's the same as 3 ÷ 20. But 20 goes into 3 zero times, right? So I need to add a decimal point and some zeros to make this division work. Let me write it out like long division.

So, 20 goes into 30 once because 20 times 1 is 20. Subtract 20 from 30, and I get 10. Then bring down the next zero, making it 100. Wait, but there are no more digits in the original number, so I can keep adding zeros. So 20 goes into 100 five times because 20 times 5 is 100. Subtract 100 from 100, and there's nothing left. So that would make the decimal 0.15?

Wait, let me check that again. Let me do the division step by step. 3 divided by 20. Since 20 is larger than 3, we write it as 0. and then proceed. Multiply numerator and denominator by 5 to make the denominator 100? Wait, that might be another way. If I multiply both numerator and denominator by 5, then 20 times 5 is 100, and 3 times 5 is 15. So 3/20 is equivalent to 15/100, which is 0.15. Oh, that's easier! So that gives me 0.15 directly.

But let me verify with the long division method to make sure. So starting with 3 divided by 20.

First, 20 goes into 3 zero times. Add a decimal point and a zero, making it 30. 20 goes into 30 once (1), multiply 20 by 1 is 20, subtract from 30, remainder 10. Bring down another zero, making it 100. 20 goes into 100 five times (5), multiply 20 by 5 is 100, subtract, remainder zero. So the result is 0.15. Yep, that's the same as before.

So both methods give me 0.15. Let me check with another approach. Maybe converting the fraction to a decimal by recognizing common denominators. Since 20 is a factor of 100, as I did earlier, multiplying numerator and denominator by 5 gives me 15/100, which is 0.15.

Alternatively, I can think of 3 divided by 20 as 3 divided by (2 times 10), which is (3 divided by 2) divided by 10. Let's see. 3 divided by 2 is 1.5, then divide by 10 is 0.15. That also works.

So all methods lead me to 0.15. Let me just confirm with a calculator to be absolutely sure. Hmm, if I do 3 divided by 20, that equals... 0.15. Yep, that's correct.

I think that's solid. So the decimal representation of 3/20 is 0.15.

**Final Answer**
The decimal representation of $\frac{3}{20}$ is \boxed{0.15}.

To convert the fraction $\frac{3}{20}$ into a decimal, we can use different approaches, all of which ultimately rely on the fundamental idea of expressing a fraction as a division of its numerator by its denominator. Here, this means calculating:

$$
\frac{3}{20} = 3 \div 20
$$

---

### Step 1: Direct Division (Long Division)

We perform the long division of 3 by 20:

- Since 20 is larger than 3, we place a decimal point and add a zero, making it 30.
- 20 goes into 30 once (1), so we write 1 after the decimal point.
- Multiply 20 by 1 to get 20, subtract from 30, getting a remainder of 10.
- Bring down the next zero to make it 100.
- 20 goes into 100 exactly five times (5), with no remainder.
- So the result is $0.15$.

---

### Step 2: Alternative Method – Adjusting the Denominator

We can also convert the fraction by multiplying both the numerator and the denominator by a number to make the denominator a power of 10:

$$
\frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100}
$$

Now, since $\frac{15}{100}$ is the same as 0.15, we have our answer.

---

### Step 3: Another Perspective – Breaking Down the Division

We can write:
$$
\frac{3}{20} = \frac{3}{2 \times 10} = \left(\frac{3}{2} \div 10\right)
$$

- $3 \div 2 = 1.5$
- $1.5 \div 10 = 0.15$

---

### Final Conclusion

All methods consistently lead to the same decimal value. Therefore, the decimal representation of $\frac{3}{20}$ is:

$$
\boxed{0.15}
$$

[Assistant]
Explanation:
Okay, so I need to convert the fraction 3/20 into a decimal. Hmm, let me think about how to do that. I remember that converting fractions to decimals usually involves division. So maybe I need to divide 3 by 20? Let me try that.

Alright, let's set it up. If I divide 3 by 20, that's the same as 3 ÷ 20. But 20 goes into 3 zero times, right? So I need to add a decimal point and some zeros to make this division work. Let me write it out like long division.

So, 20 goes into 30 once because 20 times 1 is 20. Subtract 20 from 30, and I get 10. Then bring down the next zero, making it 100. Wait, but there are no more digits in the original number, so I can keep adding zeros. So 20 goes into 100 five times because 20 times 5 is 100. Subtract 100 from 100, and there's nothing left. So that would make the decimal 0.15?

Wait, let me check that again. Let me do the division step by step. 3 divided by 20. Since 20 is larger than 3, we write it as 0. and then proceed. Multiply numerator and denominator by 5 to make the denominator 100? Wait, that might be another way. If I multiply both numerator and denominator by 5, then 20 times 5 is 100, and 3 times 5 is 15. So 3/20 is equivalent to 15/100, which is 0.15. Oh, that's easier! So that gives me 0.15 directly.

But let me verify with the long division method to make sure. So starting with 3 divided by 20.

First, 20 goes into 3 zero times. Add a decimal point and a zero, making it 30. 20 goes into 30 once (1), multiply 20 by 1 is 20, subtract from 30, remainder 10. Bring down another zero, making it 100. 20 goes into 100 five times (5), multiply 20 by 5 is 100, subtract, remainder zero. So the result is 0.15. Yep, that's the same as before.

So both methods give me 0.15. Let me check with another approach. Maybe converting the fraction to a decimal by recognizing common denominators. Since 20 is a factor of 100, as I did earlier, multiplying numerator and denominator by 5 gives me 15/100, which is 0.15.

Alternatively, I can think of 3 divided by 20 as 3 divided by (2 times 10), which is (3 divided by 2) divided by 10. Let's see. 3 divided by 2 is 1.5, then divide by 10 is 0.15. That also works.

So all methods lead me to 0.15. Let me just confirm with a calculator to be absolutely sure. Hmm, if I do 3 divided by 20, that equals... 0.15. Yep, that's correct.

I think that's solid. So the decimal representation of 3/20 is 0.15.

**Final Answer**
The decimal representation of $\frac{3}{20}$ is \boxed{0.15}.

To convert the fraction $\frac{3}{20}$ into a decimal, we can use different approaches, all of which ultimately rely on the fundamental idea of expressing a fraction as a division of its numerator by its denominator. Here, this means calculating:

$$
\frac{3}{20} = 3 \div 20
$$

---

### Step 1: Direct Division (Long Division)

We perform the long division of 3 by 20:

- Since 20 is larger than 3, we place a decimal point and add a zero, making it 30.
- 20 goes into 30 once (1), so we write 1 after the decimal point.
- Multiply 20 by 1 to get 20, subtract from 30, getting a remainder of 10.
- Bring down the next zero to make it 100.
- 20 goes into 100 exactly five times (5), with no remainder.
- So the result is $0.15$.

---

### Step 2: Alternative Method – Adjusting the Denominator

We can also convert the fraction by multiplying both the numerator and the denominator by a number to make the denominator a power of 10:

$$
\frac{3}{20} = \frac{3 \times 5}{20 \times 5} = \frac{15}{100}
$$

Now, since $\frac{15}{100}$ is the same as 0.15, we have our answer.

---

### Step 3: Another Perspective – Breaking Down the Division

We can write:
$$
\frac{3}{20} = \frac{3}{2 \times 10} = \left(\frac{3}{2} \div 10\right)
$$

- $3 \div 2 = 1.5$
- $1.5 \div 10 = 0.15$

---

### Final Conclusion

All methods consistently lead to the same decimal value. Therefore, the decimal representation of $\frac{3}{20}$ is:

$$
\boxed{0.15}
$$

Solution: Multiplying numerator and denominator by 5 gives $3/20=15/100 = \boxed{0.15}$.


Explanation:
Okay, so Ringo needs to calculate the modulus of (1 - i) raised to the 8th power. Hmm, let me think. First, I remember that the modulus of a complex number a + bi is sqrt(a² + b²). So, maybe I should start by finding the modulus of (1 - i) first, and then raise it to the 8th power? Wait, is there another way? Because when you raise a complex number to a power, there's also De Moivre's theorem, which involves the modulus and the argument. Maybe I can use that?

Let me try both approaches and see if they match. Let's start with the modulus. The complex number is 1 - i. So, a is 1 and b is -1. Then the modulus is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So |1 - i| is sqrt(2). Then, if I raise that modulus to the 8th power, it would be (sqrt(2))⁸. Let me calculate that. Since sqrt(2) is 2^(1/2), so raising that to the 8th power would be 2^(1/2 * 8) = 2^4 = 16. So that gives me 16.

But let me check with De Moivre's theorem to make sure. First, convert 1 - i to polar form. The modulus is sqrt(2) as before. Then the argument θ. Since the complex number is in the fourth quadrant (real part positive, imaginary part negative), the argument is -π/4 or 7π/4. So, in polar form, it's sqrt(2)*(cos(-π/4) + i sin(-π/4)). Then, raising this to the 8th power would be [sqrt(2)]^8 * [cos(-π/4 * 8) + i sin(-π/4 * 8)]. Let me compute each part.

First, [sqrt(2)]^8 is the same as before, which is 16. Then, the angle is -π/4 * 8 = -2π. Cos(-2π) is cos(2π) which is 1, and sin(-2π) is sin(2π) which is 0. Therefore, the result is 16*(1 + 0i) = 16. So that matches the first method. So the modulus of (1 - i)^8 is 16. Therefore, Ringo should have found 16.

But wait, let me make sure I didn't skip any steps. Alternatively, could I compute (1 - i)^8 directly by expanding? Although that might be cumbersome, but let's try for verification. Let's compute (1 - i)^2 first. (1 - i)^2 = 1^2 - 2*1*i + i^2 = 1 - 2i + (-1) = 0 - 2i = -2i. Then, (1 - i)^4 would be [ (1 - i)^2 ]^2 = (-2i)^2 = 4i² = 4*(-1) = -4. Then, (1 - i)^8 = [ (1 - i)^4 ]^2 = (-4)^2 = 16. Oh, that's straightforward! So that method also gives 16.

So all three methods: modulus raised to the power, De Moivre's theorem, and direct exponentiation, give the same result. Therefore, Ringo should have found 16.

Wait, but just to be thorough, let me check the direct calculation again. Starting with (1 - i)^2 = -2i. Then, squaring that gives (-2i)^2 = 4i² = -4. Then, squaring that again gives (-4)^2 = 16. Yep, that's correct.

Alternatively, maybe I can think about the modulus of (1 - i)^8 as |1 - i|^8, which is (sqrt(2))^8, which is 2^(4) = 16. So that's consistent.

Therefore, regardless of the method, the answer is 16. I think that's solid. No mistakes in my calculations. All methods confirm that the modulus is 16.

So, Ringo's answer should be 16.

**Final Answer**
\boxed{16}

To solve the expression $ |(1 - i)^8| $, we can use properties of complex numbers and modulus of powers.

---

### Step 1: Understand the modulus of a complex number

Given a complex number $ z = a + bi $, the modulus is defined as:

$$
|z| = \sqrt{a^2 + b^2}
$$

For $ z = 1 - i $, we have $ a = 1 $, $ b = -1 $, so:

$$
|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
$$

---

### Step 2: Use the property $ |z^n| = |z|^n $

We want $ |(1 - i)^8| $. Using the rule:

$$
|(1 - i)^8| = |1 - i|^8 = (\sqrt{2})^8
$$

Now simplify:

$$
(\sqrt{2})^8 = (2^{1/2})^8 = 2^{(1/2) \cdot 8} = 2^4 = 16
$$

---

### Step 3: Optional Verification via De Moivre’s Theorem

We can further verify using polar form. The complex number $ 1 - i $ has modulus $ \sqrt{2} $ and argument $ \theta = -\frac{\pi}{4} $ (or equivalently $ \frac{7\pi}{4} $).

Using De Moivre’s Theorem:

$$
(1 - i)^8 = ( \sqrt{2} )^8 \left[ \cos\left(-\frac{8\pi}{4}\right) + i \sin\left(-\frac{8\pi}{4}\right) \right]
$$

$$
= 16 \left[ \cos(-2\pi) + i \sin(-2\pi) \right] = 16 (1 + 0i) = 16
$$

So, the modulus is again $ |(1 - i)^8| = 16 $.

---

### Step 4: Direct Expansion (Optional, for thoroughness)

We can compute $ (1 - i)^2 $ first:

$$
(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i
$$

Then:

$$
(1 - i)^4 = (-2i)^2 = -4
$$

$$
(1 - i)^8 = (-4)^2 = 16
$$

So again, the result is consistent.

---

### Final Answer

$$
\boxed{16}
$$

[Assistant]
Explanation:
Okay, so Ringo needs to calculate the modulus of (1 - i) raised to the 8th power. Hmm, let me think. First, I remember that the modulus of a complex number a + bi is sqrt(a² + b²). So, maybe I should start by finding the modulus of (1 - i) first, and then raise it to the 8th power? Wait, is there another way? Because when you raise a complex number to a power, there's also De Moivre's theorem, which involves the modulus and the argument. Maybe I can use that?

Let me try both approaches and see if they match. Let's start with the modulus. The complex number is 1 - i. So, a is 1 and b is -1. Then the modulus is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So |1 - i| is sqrt(2). Then, if I raise that modulus to the 8th power, it would be (sqrt(2))⁸. Let me calculate that. Since sqrt(2) is 2^(1/2), so raising that to the 8th power would be 2^(1/2 * 8) = 2^4 = 16. So that gives me 16.

But let me check with De Moivre's theorem to make sure. First, convert 1 - i to polar form. The modulus is sqrt(2) as before. Then the argument θ. Since the complex number is in the fourth quadrant (real part positive, imaginary part negative), the argument is -π/4 or 7π/4. So, in polar form, it's sqrt(2)*(cos(-π/4) + i sin(-π/4)). Then, raising this to the 8th power would be [sqrt(2)]^8 * [cos(-π/4 * 8) + i sin(-π/4 * 8)]. Let me compute each part.

First, [sqrt(2)]^8 is the same as before, which is 16. Then, the angle is -π/4 * 8 = -2π. Cos(-2π) is cos(2π) which is 1, and sin(-2π) is sin(2π) which is 0. Therefore, the result is 16*(1 + 0i) = 16. So that matches the first method. So the modulus of (1 - i)^8 is 16. Therefore, Ringo should have found 16.

But wait, let me make sure I didn't skip any steps. Alternatively, could I compute (1 - i)^8 directly by expanding? Although that might be cumbersome, but let's try for verification. Let's compute (1 - i)^2 first. (1 - i)^2 = 1^2 - 2*1*i + i^2 = 1 - 2i + (-1) = 0 - 2i = -2i. Then, (1 - i)^4 would be [ (1 - i)^2 ]^2 = (-2i)^2 = 4i² = 4*(-1) = -4. Then, (1 - i)^8 = [ (1 - i)^4 ]^2 = (-4)^2 = 16. Oh, that's straightforward! So that method also gives 16.

So all three methods: modulus raised to the power, De Moivre's theorem, and direct exponentiation, give the same result. Therefore, Ringo should have found 16.

Wait, but just to be thorough, let me check the direct calculation again. Starting with (1 - i)^2 = -2i. Then, squaring that gives (-2i)^2 = 4i² = -4. Then, squaring that again gives (-4)^2 = 16. Yep, that's correct.

Alternatively, maybe I can think about the modulus of (1 - i)^8 as |1 - i|^8, which is (sqrt(2))^8, which is 2^(4) = 16. So that's consistent.

Therefore, regardless of the method, the answer is 16. I think that's solid. No mistakes in my calculations. All methods confirm that the modulus is 16.

So, Ringo's answer should be 16.

**Final Answer**
\boxed{16}

To solve the expression $ |(1 - i)^8| $, we can use properties of complex numbers and modulus of powers.

---

### Step 1: Understand the modulus of a complex number

Given a complex number $ z = a + bi $, the modulus is defined as:

$$
|z| = \sqrt{a^2 + b^2}
$$

For $ z = 1 - i $, we have $ a = 1 $, $ b = -1 $, so:

$$
|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
$$

---

### Step 2: Use the property $ |z^n| = |z|^n $

We want $ |(1 - i)^8| $. Using the rule:

$$
|(1 - i)^8| = |1 - i|^8 = (\sqrt{2})^8
$$

Now simplify:

$$
(\sqrt{2})^8 = (2^{1/2})^8 = 2^{(1/2) \cdot 8} = 2^4 = 16
$$

---

### Step 3: Optional Verification via De Moivre’s Theorem

We can further verify using polar form. The complex number $ 1 - i $ has modulus $ \sqrt{2} $ and argument $ \theta = -\frac{\pi}{4} $ (or equivalently $ \frac{7\pi}{4} $).

Using De Moivre’s Theorem:

$$
(1 - i)^8 = ( \sqrt{2} )^8 \left[ \cos\left(-\frac{8\pi}{4}\right) + i \sin\left(-\frac{8\pi}{4}\right) \right]
$$

$$
= 16 \left[ \cos(-2\pi) + i \sin(-2\pi) \right] = 16 (1 + 0i) = 16
$$

So, the modulus is again $ |(1 - i)^8| = 16 $.

---

### Step 4: Direct Expansion (Optional, for thoroughness)

We can compute $ (1 - i)^2 $ first:

$$
(1 - i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i
$$

Then:

$$
(1 - i)^4 = (-2i)^2 = -4
$$

$$
(1 - i)^8 = (-4)^2 = 16
$$

So again, the result is consistent.

---

### Final Answer

$$
\boxed{16}
$$

Solution: We know that magnitudes of complex numbers are multiplicative: the magnitude of $|ab|$ is the product $|a|\cdot |b|$. Thus, \[\left|\left(1-i\right)^8\right|=\left|1-i\right|^8\]The magnitude of $1-i$ is $\sqrt{1^2+(-1)^2}=\sqrt{2}$; thus our answer is $\left(\sqrt{2}\right) ^8=\boxed{16}$. Ringo didn't give much of a reprieve.