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 start by recalling what roots of unity are. A root of unity is a complex number that satisfies zⁿ = 1 for some positive integer n. So, essentially, these are the solutions to the equation zⁿ = 1. The roots of unity are located on the unit circle in the complex plane and are equally spaced around the circle.

Now, the problem is about the roots of the polynomial z⁴ + z² + 1 = 0. I need to find the smallest n such that each of these roots is an nth root of unity. That means each root should satisfy zⁿ = 1. So, first, maybe I should find the roots of the given polynomial and then determine what n would make them roots of unity.

But solving z⁴ + z² + 1 = 0 directly might be a bit tricky. Let me think. Maybe I can factor this polynomial or relate it to a known equation. Let me try substituting w = z². Then the equation becomes w² + w + 1 = 0. That seems simpler. Let me solve for w first.

So, solving w² + w + 1 = 0. Using the quadratic formula: w = [-1 ± √(1 - 4)] / 2 = [-1 ± √(-3)] / 2 = (-1 ± i√3)/2. Therefore, the solutions for w are complex numbers. So, going back to z, since w = z², we have z² = (-1 ± i√3)/2. Therefore, each of these w's is a complex number. Let me recall that (-1 ± i√3)/2 are the primitive third roots of unity. Wait, because the cube roots of unity are 1, (-1 + i√3)/2, and (-1 - i√3)/2. So, these are the non-real cube roots of unity. Therefore, w is a primitive 3rd root of unity. Therefore, z² is a primitive 3rd root of unity. Therefore, z is a square root of a primitive 3rd root of unity.

So, if I can find the roots of z² = ω, where ω is a primitive 3rd root of unity, then those roots would be... Let me think. Let me denote ω = e^(2πi/3) = -1/2 + i√3/2. Similarly, the other root is ω² = e^(4πi/3) = -1/2 - i√3/2. So, solving z² = ω. Let me find the square roots of ω. Let me recall that if I have a complex number in polar form, then taking roots is straightforward. Let me write ω in polar form. Since ω is e^(2πi/3), so its magnitude is 1, and its angle is 2π/3. Therefore, the square roots of ω would have magnitude 1 and angles (2π/3 + 2πk)/2 for k = 0, 1. Therefore, angles π/3 and π/3 + π = 4π/3. Therefore, the square roots of ω are e^(iπ/3) and e^(i4π/3). Similarly, the square roots of ω² would be e^(i2π/3) and e^(i5π/3). Therefore, the roots of z⁴ + z² + 1 = 0 are e^(iπ/3), e^(i4π/3), e^(i2π/3), and e^(i5π/3). Wait, but let me check that.

Wait, original equation: z⁴ + z² + 1 = 0. Let me check if these roots satisfy the equation. Let me take z = e^(iπ/3). Let me compute z²: e^(i2π/3) which is ω. Then z⁴ is (z²)² = ω². Therefore, z⁴ + z² + 1 = ω² + ω + 1. But ω² + ω + 1 = 0, since ω is a root of w² + w + 1 = 0. Therefore, yes, that works. Similarly for the other roots. Therefore, the four roots of the equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), and e^(i5π/3). Wait, but hold on, e^(iπ/3) is 60 degrees, e^(i2π/3) is 120 degrees, e^(i4π/3) is 240 degrees, and e^(i5π/3) is 300 degrees. So, these are the 6th roots of unity except for 1 and -1? Wait, the 6th roots of unity are e^(iπk/3) for k = 0, 1, 2, 3, 4, 5. So, e^(iπ/3), e^(i2π/3), e^(iπ) = -1, e^(i4π/3), e^(i5π/3), and e^(i6π/3) = e^(i2π) = 1. Therefore, the roots of our equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). So, they are the primitive 6th roots of unity? Wait, but e^(iπ/3) is a primitive 6th root of unity because its order is 6. Let me check. The order of e^(iπ/3) is the smallest positive integer k such that (e^(iπ/3))^k = 1. So, e^(iπk/3) = 1 implies πk/3 = 2πm 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, e^(i2π/3) is a primitive 3rd root of unity? Wait, let me check. The order of e^(i2π/3) is the smallest k such that e^(i2πk/3) = 1. So, 2πk/3 = 2πm => k/3 = m => k = 3m. 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, since 4π/3 is equivalent to -2π/3, and same order. Similarly, e^(i5π/3) is a primitive 6th root of unity. Therefore, the roots of the equation z⁴ + z² + 1 = 0 are two primitive 6th roots of unity and two primitive 3rd roots of unity. Therefore, if we want all of these roots to be nth roots of unity, then n must be a common multiple of 6 and 3. The least common multiple of 6 and 3 is 6. Wait, but hold on. Let me check. If n is 6, then all the roots are 6th roots of unity. But the primitive 3rd roots of unity are also 6th roots of unity, since 3 divides 6. Therefore, yes, all roots are 6th roots of unity. Therefore, n=6? But wait, let me check if maybe there is a smaller n. Wait, the roots are 6th roots of unity, but maybe they are also roots of unity of lower order? For example, the primitive 3rd roots of unity are 3rd roots of unity, but they are also 6th roots of unity. So, if n is 6, then they are included. However, if we take n=3, then the roots e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity. Because e^(iπ/3) raised to the 3rd power is e^(iπ) = -1 ≠ 1. Therefore, they are not 3rd roots of unity. Similarly, n=4? Let me check. Let me check if the roots are 4th roots of unity. The 4th roots of unity are 1, i, -1, -i. Our roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). None of these are i or -i or 1 or -1. So, n=4 is out. How about n=6? Let me check. The 6th roots of unity are e^(iπk/3) for k=0,...,5. Which includes our roots. Therefore, n=6 works. But is there a smaller n? Let me check n=12? No, 6 is smaller. Wait, but maybe n= something else? Let me think. Let me check if maybe n=12? Wait, but 6 is smaller. Wait, but maybe n= something else. Let me think again. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Let me check their orders. The order of e^(iπ/3) is 6, as we saw. The order of e^(i2π/3) is 3. Therefore, for all roots to be nth roots of unity, n must be a multiple of 6 and 3. The least common multiple of 6 and 3 is 6. Therefore, n=6. Therefore, the minimal n is 6. Wait, but let me check if maybe there is a smaller n. For example, n=12? No, 6 is smaller. Wait, but maybe n= something else? Let me think. Suppose n= 12. Then all roots would be 12th roots of unity, but since 6 divides 12, they are also 6th roots of unity. But since 6 is smaller, n=6 is better. So, is there a smaller n than 6? Let me check n= 3. As before, the roots e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity. Similarly, n=4, n=5? Let me check n=5. The 5th roots of unity are e^(2πik/5). Our roots are e^(πi/3) = e^(2πi/6), e^(2πi/3), e^(4πi/3), e^(5πi/3). So, none of these are 5th roots of unity. For example, e^(2πi/6) is not a 5th root of unity. Similarly, n=2? No. n=1? No. So, seems like n=6 is the minimal. Wait, but let me check if maybe there is a different approach. Let me think about the original equation. Maybe I can relate z⁴ + z² + 1 = 0 to cyclotomic polynomials. Because cyclotomic polynomials are minimal polynomials over integers for primitive roots of unity. Let me recall that the cyclotomic polynomial Φ_n(z) is the minimal polynomial whose roots are the primitive nth roots of unity. So, maybe z⁴ + z² + 1 is a cyclotomic polynomial? Let me check. Let me recall that Φ_3(z) = z² + z + 1. Φ_6(z) is z² - z + 1. Wait, but our polynomial is z⁴ + z² + 1. Let me see. Let me note that z⁶ - 1 = (z³ - 1)(z³ + 1) = (z - 1)(z² + z + 1)(z + 1)(z² - z + 1). Therefore, z⁶ - 1 factors into Φ_1(z)Φ_3(z)Φ_2(z)Φ_6(z). Wait, but z⁴ + z² + 1. Let me see. Let me note that z⁶ - 1 = (z²)³ - 1 = (z² - 1)(z⁴ + z² + 1). Therefore, z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1). Therefore, z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1) = Φ_3(z)Φ_6(z). Wait, because z⁶ - 1 factors as Φ_1(z)Φ_2(z)Φ_3(z)Φ_6(z). Therefore, dividing by z² - 1, which is Φ_1(z)Φ_2(z), we get Φ_3(z)Φ_6(z). Therefore, z⁴ + z² + 1 = Φ_3(z)Φ_6(z). Therefore, the roots of z⁴ + z² + 1 = 0 are the roots of Φ_3(z) and Φ_6(z). Therefore, the roots are the primitive 3rd roots of unity and the primitive 6th roots of unity. Therefore, the roots are the primitive 3rd and 6th roots of unity. Therefore, to have all roots be nth roots of unity, n must be a multiple of both 3 and 6. The least common multiple of 3 and 6 is 6. Therefore, n=6. Therefore, the minimal n is 6. Therefore, the answer is 6. Wait, but let me check again. Let me confirm that Φ_3(z)Φ_6(z) is indeed z⁴ + z² + 1. Let me compute Φ_3(z) = z² + z + 1, Φ_6(z) = z² - z + 1. Multiplying them: (z² + z + 1)(z² - z + 1) = z⁴ - z³ + z² + z³ - z² + z + z² - z + 1. Let me compute term by term:

First, z² * z² = z⁴.

z² * (-z) = -z³.

z² * 1 = z².

z * z² = z³.

z * (-z) = -z².

z * 1 = z.

1 * z² = z².

1 * (-z) = -z.

1 * 1 = 1.

Now, combining like terms:

z⁴ + (-z³ + z³) + (z² - z² + z²) + (z - z) + 1.

Simplify:

z⁴ + 0 + z² + 0 + 1 = z⁴ + z² + 1. Yes! Perfect. Therefore, the polynomial factors into Φ_3(z)Φ_6(z). Therefore, the roots are the primitive 3rd roots of unity and the primitive 6th roots of unity. Therefore, the roots are the 3rd and 6th roots of unity, but not the 1st or 2nd. Therefore, to have all roots be nth roots of unity, n must be a multiple of both 3 and 6. The least common multiple of 3 and 6 is 6. Therefore, n=6. Therefore, the minimal n is 6. Therefore, the answer is 6. Wait, but let me check once more. Let me check if all roots of z⁴ + z² + 1 are indeed 6th roots of unity. The roots are e^(2πi/3), e^(4πi/3), e^(πi/3), e^(5πi/3). Let me check if they are 6th roots of unity. For example, e^(πi/3) is e^(2πi/6), which is a primitive 6th root of unity. Similarly, e^(5πi/3) is e^(10πi/6) = e^(2πi*(5/6)), which is also a primitive 6th root of unity. The roots e^(2πi/3) and e^(4πi/3) are e^(4πi/6) and e^(8πi/6) = e^(4πi/3). Wait, 2π/3 is 4π/6. So, they are e^(4πi/6) and e^(8πi/6). But 4π/6 reduces to 2π/3, which is a primitive 3rd root of unity. However, since 3 divides 6, they are also 6th roots of unity. Therefore, all roots are 6th roots of unity. Therefore, n=6 works. Is there a smaller n? Let me check n=6. Suppose n=6. Then, yes, all roots are 6th roots of unity. If n=3, then the roots e^(πi/3) and e^(5πi/3) are not 3rd roots of unity. If n=4, they are not 4th roots. If n=5, they are not 5th roots. If n=2, no. n=1, no. Therefore, n=6 is indeed the minimal. Therefore, the answer is 6. Wait, but let me check if maybe there is a different n. For example, suppose n=12. Then, all roots are 12th roots of unity, but since 6 divides 12, n=6 is smaller. Therefore, 6 is the minimal. Therefore, I think the answer is 6. However, let me check once again. Let me think about the orders of the roots. The roots are of order 3 and 6. Therefore, n must be a multiple of 6. Therefore, the minimal n is 6. Therefore, the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine the smallest positive integer $ n $ such that all the roots of the polynomial
$$
z^4 + z^2 + 1 = 0
$$
are $ n $th roots of unity, we begin by analyzing the structure of the roots.

---

### Step 1: Factor the Polynomial

Let us make a substitution to simplify the equation. Let $ w = z^2 $. Then the equation becomes:
$$
w^2 + w + 1 = 0
$$
This is a quadratic equation with complex roots:
$$
w = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
$$
These are the **primitive 3rd roots of unity**, denoted $ \omega $ and $ \omega^2 $, where $ \omega = e^{2\pi i / 3} $.

Now, recall that $ w = z^2 $, so the original equation becomes:
$$
z^2 = \omega \quad \text{or} \quad z^2 = \omega^2
$$
Taking square roots of both sides, we find the four roots of the original equation:
$$
z = \pm \sqrt{\omega}, \quad \pm \sqrt{\omega^2}
$$
These are the **four roots** of the equation $ z^4 + z^2 + 1 = 0 $.

---

### Step 2: Identify the Roots

Let’s write the roots in exponential form. Since $ \omega = e^{2\pi i / 3} $, we have:
- $ \sqrt{\omega} = e^{\pi i / 3} $ and $ e^{4\pi i / 3} $
- $ \sqrt{\omega^2} = e^{2\pi i / 3} $ and $ e^{5\pi i / 3} $

So the four roots are:
$$
e^{\pi i / 3}, \quad e^{2\pi i / 3}, \quad e^{4\pi i / 3}, \quad e^{5\pi i / 3}
$$

These are the **primitive 6th roots of unity** and the **primitive 3rd roots of unity**. Specifically:
- $ e^{\pi i / 3} $ and $ e^{5\pi i / 3} $ are **primitive 6th roots of unity**
- $ e^{2\pi i / 3} $ and $ e^{4\pi i / 3} $ are **primitive 3rd roots of unity**

---

### Step 3: Determine the Smallest $ n $

We are looking for the **smallest** positive integer $ n $ such that **all** these roots are $ n $th roots of unity.

- The **primitive 3rd roots of unity** are 3rd roots of unity.
- The **primitive 6th roots of unity** are 6th roots of unity.

Thus, for all roots to be $ n $th roots of unity, $ n $ must be a **common multiple** of both 3 and 6.

The **least common multiple** of 3 and 6 is:
$$
\text{lcm}(3, 6) = 6
$$

Therefore, the **smallest** such $ n $ is:
$$
\boxed{6}
$$

---

### 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 start by recalling what roots of unity are. A root of unity is a complex number that satisfies zⁿ = 1 for some positive integer n. So, essentially, these are the solutions to the equation zⁿ = 1. The roots of unity are located on the unit circle in the complex plane and are equally spaced around the circle.

Now, the problem is about the roots of the polynomial z⁴ + z² + 1 = 0. I need to find the smallest n such that each of these roots is an nth root of unity. That means each root should satisfy zⁿ = 1. So, first, maybe I should find the roots of the given polynomial and then determine what n would make them roots of unity.

But solving z⁴ + z² + 1 = 0 directly might be a bit tricky. Let me think. Maybe I can factor this polynomial or relate it to a known equation. Let me try substituting w = z². Then the equation becomes w² + w + 1 = 0. That seems simpler. Let me solve for w first.

So, solving w² + w + 1 = 0. Using the quadratic formula: w = [-1 ± √(1 - 4)] / 2 = [-1 ± √(-3)] / 2 = (-1 ± i√3)/2. Therefore, the solutions for w are complex numbers. So, going back to z, since w = z², we have z² = (-1 ± i√3)/2. Therefore, each of these w's is a complex number. Let me recall that (-1 ± i√3)/2 are the primitive third roots of unity. Wait, because the cube roots of unity are 1, (-1 + i√3)/2, and (-1 - i√3)/2. So, these are the non-real cube roots of unity. Therefore, w is a primitive 3rd root of unity. Therefore, z² is a primitive 3rd root of unity. Therefore, z is a square root of a primitive 3rd root of unity.

So, if I can find the roots of z² = ω, where ω is a primitive 3rd root of unity, then those roots would be... Let me think. Let me denote ω = e^(2πi/3) = -1/2 + i√3/2. Similarly, the other root is ω² = e^(4πi/3) = -1/2 - i√3/2. So, solving z² = ω. Let me find the square roots of ω. Let me recall that if I have a complex number in polar form, then taking roots is straightforward. Let me write ω in polar form. Since ω is e^(2πi/3), so its magnitude is 1, and its angle is 2π/3. Therefore, the square roots of ω would have magnitude 1 and angles (2π/3 + 2πk)/2 for k = 0, 1. Therefore, angles π/3 and π/3 + π = 4π/3. Therefore, the square roots of ω are e^(iπ/3) and e^(i4π/3). Similarly, the square roots of ω² would be e^(i2π/3) and e^(i5π/3). Therefore, the roots of z⁴ + z² + 1 = 0 are e^(iπ/3), e^(i4π/3), e^(i2π/3), and e^(i5π/3). Wait, but let me check that.

Wait, original equation: z⁴ + z² + 1 = 0. Let me check if these roots satisfy the equation. Let me take z = e^(iπ/3). Let me compute z²: e^(i2π/3) which is ω. Then z⁴ is (z²)² = ω². Therefore, z⁴ + z² + 1 = ω² + ω + 1. But ω² + ω + 1 = 0, since ω is a root of w² + w + 1 = 0. Therefore, yes, that works. Similarly for the other roots. Therefore, the four roots of the equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), and e^(i5π/3). Wait, but hold on, e^(iπ/3) is 60 degrees, e^(i2π/3) is 120 degrees, e^(i4π/3) is 240 degrees, and e^(i5π/3) is 300 degrees. So, these are the 6th roots of unity except for 1 and -1? Wait, the 6th roots of unity are e^(iπk/3) for k = 0, 1, 2, 3, 4, 5. So, e^(iπ/3), e^(i2π/3), e^(iπ) = -1, e^(i4π/3), e^(i5π/3), and e^(i6π/3) = e^(i2π) = 1. Therefore, the roots of our equation are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). So, they are the primitive 6th roots of unity? Wait, but e^(iπ/3) is a primitive 6th root of unity because its order is 6. Let me check. The order of e^(iπ/3) is the smallest positive integer k such that (e^(iπ/3))^k = 1. So, e^(iπk/3) = 1 implies πk/3 = 2πm 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, e^(i2π/3) is a primitive 3rd root of unity? Wait, let me check. The order of e^(i2π/3) is the smallest k such that e^(i2πk/3) = 1. So, 2πk/3 = 2πm => k/3 = m => k = 3m. 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, since 4π/3 is equivalent to -2π/3, and same order. Similarly, e^(i5π/3) is a primitive 6th root of unity. Therefore, the roots of the equation z⁴ + z² + 1 = 0 are two primitive 6th roots of unity and two primitive 3rd roots of unity. Therefore, if we want all of these roots to be nth roots of unity, then n must be a common multiple of 6 and 3. The least common multiple of 6 and 3 is 6. Wait, but hold on. Let me check. If n is 6, then all the roots are 6th roots of unity. But the primitive 3rd roots of unity are also 6th roots of unity, since 3 divides 6. Therefore, yes, all roots are 6th roots of unity. Therefore, n=6? But wait, let me check if maybe there is a smaller n. Wait, the roots are 6th roots of unity, but maybe they are also roots of unity of lower order? For example, the primitive 3rd roots of unity are 3rd roots of unity, but they are also 6th roots of unity. So, if n is 6, then they are included. However, if we take n=3, then the roots e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity. Because e^(iπ/3) raised to the 3rd power is e^(iπ) = -1 ≠ 1. Therefore, they are not 3rd roots of unity. Similarly, n=4? Let me check. Let me check if the roots are 4th roots of unity. The 4th roots of unity are 1, i, -1, -i. Our roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). None of these are i or -i or 1 or -1. So, n=4 is out. How about n=6? Let me check. The 6th roots of unity are e^(iπk/3) for k=0,...,5. Which includes our roots. Therefore, n=6 works. But is there a smaller n? Let me check n=12? No, 6 is smaller. Wait, but maybe n= something else? Let me think. Let me check if maybe n=12? Wait, but 6 is smaller. Wait, but maybe n= something else. Let me think again. The roots are e^(iπ/3), e^(i2π/3), e^(i4π/3), e^(i5π/3). Let me check their orders. The order of e^(iπ/3) is 6, as we saw. The order of e^(i2π/3) is 3. Therefore, for all roots to be nth roots of unity, n must be a multiple of 6 and 3. The least common multiple of 6 and 3 is 6. Therefore, n=6. Therefore, the minimal n is 6. Wait, but let me check if maybe there is a smaller n. For example, n=12? No, 6 is smaller. Wait, but maybe n= something else? Let me think. Suppose n= 12. Then all roots would be 12th roots of unity, but since 6 divides 12, they are also 6th roots of unity. But since 6 is smaller, n=6 is better. So, is there a smaller n than 6? Let me check n= 3. As before, the roots e^(iπ/3) and e^(i5π/3) are not 3rd roots of unity. Similarly, n=4, n=5? Let me check n=5. The 5th roots of unity are e^(2πik/5). Our roots are e^(πi/3) = e^(2πi/6), e^(2πi/3), e^(4πi/3), e^(5πi/3). So, none of these are 5th roots of unity. For example, e^(2πi/6) is not a 5th root of unity. Similarly, n=2? No. n=1? No. So, seems like n=6 is the minimal. Wait, but let me check if maybe there is a different approach. Let me think about the original equation. Maybe I can relate z⁴ + z² + 1 = 0 to cyclotomic polynomials. Because cyclotomic polynomials are minimal polynomials over integers for primitive roots of unity. Let me recall that the cyclotomic polynomial Φ_n(z) is the minimal polynomial whose roots are the primitive nth roots of unity. So, maybe z⁴ + z² + 1 is a cyclotomic polynomial? Let me check. Let me recall that Φ_3(z) = z² + z + 1. Φ_6(z) is z² - z + 1. Wait, but our polynomial is z⁴ + z² + 1. Let me see. Let me note that z⁶ - 1 = (z³ - 1)(z³ + 1) = (z - 1)(z² + z + 1)(z + 1)(z² - z + 1). Therefore, z⁶ - 1 factors into Φ_1(z)Φ_3(z)Φ_2(z)Φ_6(z). Wait, but z⁴ + z² + 1. Let me see. Let me note that z⁶ - 1 = (z²)³ - 1 = (z² - 1)(z⁴ + z² + 1). Therefore, z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1). Therefore, z⁴ + z² + 1 = (z⁶ - 1)/(z² - 1) = Φ_3(z)Φ_6(z). Wait, because z⁶ - 1 factors as Φ_1(z)Φ_2(z)Φ_3(z)Φ_6(z). Therefore, dividing by z² - 1, which is Φ_1(z)Φ_2(z), we get Φ_3(z)Φ_6(z). Therefore, z⁴ + z² + 1 = Φ_3(z)Φ_6(z). Therefore, the roots of z⁴ + z² + 1 = 0 are the roots of Φ_3(z) and Φ_6(z). Therefore, the roots are the primitive 3rd roots of unity and the primitive 6th roots of unity. Therefore, the roots are the primitive 3rd and 6th roots of unity. Therefore, to have all roots be nth roots of unity, n must be a multiple of both 3 and 6. The least common multiple of 3 and 6 is 6. Therefore, n=6. Therefore, the minimal n is 6. Therefore, the answer is 6. Wait, but let me check again. Let me confirm that Φ_3(z)Φ_6(z) is indeed z⁴ + z² + 1. Let me compute Φ_3(z) = z² + z + 1, Φ_6(z) = z² - z + 1. Multiplying them: (z² + z + 1)(z² - z + 1) = z⁴ - z³ + z² + z³ - z² + z + z² - z + 1. Let me compute term by term:

First, z² * z² = z⁴.

z² * (-z) = -z³.

z² * 1 = z².

z * z² = z³.

z * (-z) = -z².

z * 1 = z.

1 * z² = z².

1 * (-z) = -z.

1 * 1 = 1.

Now, combining like terms:

z⁴ + (-z³ + z³) + (z² - z² + z²) + (z - z) + 1.

Simplify:

z⁴ + 0 + z² + 0 + 1 = z⁴ + z² + 1. Yes! Perfect. Therefore, the polynomial factors into Φ_3(z)Φ_6(z). Therefore, the roots are the primitive 3rd roots of unity and the primitive 6th roots of unity. Therefore, the roots are the 3rd and 6th roots of unity, but not the 1st or 2nd. Therefore, to have all roots be nth roots of unity, n must be a multiple of both 3 and 6. The least common multiple of 3 and 6 is 6. Therefore, n=6. Therefore, the minimal n is 6. Therefore, the answer is 6. Wait, but let me check once more. Let me check if all roots of z⁴ + z² + 1 are indeed 6th roots of unity. The roots are e^(2πi/3), e^(4πi/3), e^(πi/3), e^(5πi/3). Let me check if they are 6th roots of unity. For example, e^(πi/3) is e^(2πi/6), which is a primitive 6th root of unity. Similarly, e^(5πi/3) is e^(10πi/6) = e^(2πi*(5/6)), which is also a primitive 6th root of unity. The roots e^(2πi/3) and e^(4πi/3) are e^(4πi/6) and e^(8πi/6) = e^(4πi/3). Wait, 2π/3 is 4π/6. So, they are e^(4πi/6) and e^(8πi/6). But 4π/6 reduces to 2π/3, which is a primitive 3rd root of unity. However, since 3 divides 6, they are also 6th roots of unity. Therefore, all roots are 6th roots of unity. Therefore, n=6 works. Is there a smaller n? Let me check n=6. Suppose n=6. Then, yes, all roots are 6th roots of unity. If n=3, then the roots e^(πi/3) and e^(5πi/3) are not 3rd roots of unity. If n=4, they are not 4th roots. If n=5, they are not 5th roots. If n=2, no. n=1, no. Therefore, n=6 is indeed the minimal. Therefore, the answer is 6. Wait, but let me check if maybe there is a different n. For example, suppose n=12. Then, all roots are 12th roots of unity, but since 6 divides 12, n=6 is smaller. Therefore, 6 is the minimal. Therefore, I think the answer is 6. However, let me check once again. Let me think about the orders of the roots. The roots are of order 3 and 6. Therefore, n must be a multiple of 6. Therefore, the minimal n is 6. Therefore, the answer is 6. Therefore, I think the answer is \boxed{6}.

**Final Answer**
\boxed{6}

To determine the smallest positive integer $ n $ such that all the roots of the polynomial
$$
z^4 + z^2 + 1 = 0
$$
are $ n $th roots of unity, we begin by analyzing the structure of the roots.

---

### Step 1: Factor the Polynomial

Let us make a substitution to simplify the equation. Let $ w = z^2 $. Then the equation becomes:
$$
w^2 + w + 1 = 0
$$
This is a quadratic equation with complex roots:
$$
w = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
$$
These are the **primitive 3rd roots of unity**, denoted $ \omega $ and $ \omega^2 $, where $ \omega = e^{2\pi i / 3} $.

Now, recall that $ w = z^2 $, so the original equation becomes:
$$
z^2 = \omega \quad \text{or} \quad z^2 = \omega^2
$$
Taking square roots of both sides, we find the four roots of the original equation:
$$
z = \pm \sqrt{\omega}, \quad \pm \sqrt{\omega^2}
$$
These are the **four roots** of the equation $ z^4 + z^2 + 1 = 0 $.

---

### Step 2: Identify the Roots

Let’s write the roots in exponential form. Since $ \omega = e^{2\pi i / 3} $, we have:
- $ \sqrt{\omega} = e^{\pi i / 3} $ and $ e^{4\pi i / 3} $
- $ \sqrt{\omega^2} = e^{2\pi i / 3} $ and $ e^{5\pi i / 3} $

So the four roots are:
$$
e^{\pi i / 3}, \quad e^{2\pi i / 3}, \quad e^{4\pi i / 3}, \quad e^{5\pi i / 3}
$$

These are the **primitive 6th roots of unity** and the **primitive 3rd roots of unity**. Specifically:
- $ e^{\pi i / 3} $ and $ e^{5\pi i / 3} $ are **primitive 6th roots of unity**
- $ e^{2\pi i / 3} $ and $ e^{4\pi i / 3} $ are **primitive 3rd roots of unity**

---

### Step 3: Determine the Smallest $ n $

We are looking for the **smallest** positive integer $ n $ such that **all** these roots are $ n $th roots of unity.

- The **primitive 3rd roots of unity** are 3rd roots of unity.
- The **primitive 6th roots of unity** are 6th roots of unity.

Thus, for all roots to be $ n $th roots of unity, $ n $ must be a **common multiple** of both 3 and 6.

The **least common multiple** of 3 and 6 is:
$$
\text{lcm}(3, 6) = 6
$$

Therefore, the **smallest** such $ n $ is:
$$
\boxed{6}
$$

---

### 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 I raise 4 to that exponent, I get 8. In other words, find x where 4^x = 8. Alright, let me recall how exponents work.

First, I know that 4 is 2 squared, and 8 is 2 cubed. Maybe I can express both numbers with the same base? Let me try that. If I rewrite 4 as 2^2, then 4^x would be (2^2)^x. Using the exponent rule (a^b)^c = a^(b*c), that becomes 2^(2x). So, 2^(2x) = 8. But 8 is 2^3, right? So substituting that in, I get 2^(2x) = 2^3.

Since the bases are the same, the exponents must be equal. Therefore, 2x = 3. Solving for x, I divide both sides by 2, so x = 3/2. Let me check if that works. If I take 4^(3/2), that should be equal to 8. Let me compute 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 = 8. Yep, that checks out. So the exponent is 3/2.

Wait, let me make sure there's no other way to approach this. Maybe using logarithms? Let me try that method to confirm. Taking the logarithm of both sides of 4^x = 8. Let's use natural logarithm, but any base would work.

ln(4^x) = ln(8). Using the logarithm power rule, that becomes x * ln(4) = ln(8). Therefore, 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). So substituting back, x = (3 ln(2)) / (2 ln(2)) = 3/2. The ln(2) cancels out, leaving 3/2. So that's the same answer as before.

Another way to think about it: maybe express both numbers as powers of 2. Since 4 is 2^2 and 8 is 2^3, so 4^x = 2^(2x) = 2^3. Therefore, 2x = 3, so x = 3/2. Yep, same result.

Is there a possibility that there's another exponent? For example, negative exponents? Let me check. If x was negative, then 4^x would be 1/(4^|x|). But 8 is positive, so that's possible, but 4^x = 8 would require 4^x to be greater than 1, so x must be positive. So negative exponents are not going to work here.

What about fractional exponents? Well, we already considered that with 3/2, which is a fraction. So that's the answer.

Alternatively, maybe using common logarithms? Let me try that. log(4^x) = log(8). Then x log(4) = log(8). Therefore, x = log(8)/log(4). Again, using base 10. Let me compute that.

log(8) is log(2^3) = 3 log(2), and log(4) is log(2^2) = 2 log(2). So again, x = 3 log(2)/2 log(2) = 3/2. Same answer.

So all methods lead to x = 3/2. Therefore, the power of 4 that equals 8 is 3/2.

Let me just verify once more with actual calculation. 4^(3/2). The exponent 3/2 can be thought of as taking the square root first and then cubing, or cubing first and then taking the square root.

Square root of 4 is 2, then cubed is 8. Alternatively, 4 cubed is 64, then square root of 64 is 8. Either way, it's 8. So that's correct.

Therefore, I can be confident that the answer is 3/2.

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

To determine what power of 4 equals 8, we are essentially solving the equation:

$$
4^x = 8
$$

---

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

We know that:

- $ 4 = 2^2 $
- $ 8 = 2^3 $

So, we can rewrite the equation using base 2:

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

---

### Step 2: Apply Exponent Rules

Using the power of a power rule $ (a^b)^c = a^{bc} $, the left-hand side becomes:

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

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

$$
2x = 3
$$

---

### Step 3: Solve for $ x $

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

---

### Step 4: Verify the Solution

Let’s verify that $ 4^{3/2} = 8 $:

$$
4^{3/2} = (4^{1/2})^3 = (\sqrt{4})^3 = 2^3 = 8
$$

This confirms the solution is correct.

---

### Final Answer

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

[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 I raise 4 to that exponent, I get 8. In other words, find x where 4^x = 8. Alright, let me recall how exponents work.

First, I know that 4 is 2 squared, and 8 is 2 cubed. Maybe I can express both numbers with the same base? Let me try that. If I rewrite 4 as 2^2, then 4^x would be (2^2)^x. Using the exponent rule (a^b)^c = a^(b*c), that becomes 2^(2x). So, 2^(2x) = 8. But 8 is 2^3, right? So substituting that in, I get 2^(2x) = 2^3.

Since the bases are the same, the exponents must be equal. Therefore, 2x = 3. Solving for x, I divide both sides by 2, so x = 3/2. Let me check if that works. If I take 4^(3/2), that should be equal to 8. Let me compute 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 = 8. Yep, that checks out. So the exponent is 3/2.

Wait, let me make sure there's no other way to approach this. Maybe using logarithms? Let me try that method to confirm. Taking the logarithm of both sides of 4^x = 8. Let's use natural logarithm, but any base would work.

ln(4^x) = ln(8). Using the logarithm power rule, that becomes x * ln(4) = ln(8). Therefore, 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). So substituting back, x = (3 ln(2)) / (2 ln(2)) = 3/2. The ln(2) cancels out, leaving 3/2. So that's the same answer as before.

Another way to think about it: maybe express both numbers as powers of 2. Since 4 is 2^2 and 8 is 2^3, so 4^x = 2^(2x) = 2^3. Therefore, 2x = 3, so x = 3/2. Yep, same result.

Is there a possibility that there's another exponent? For example, negative exponents? Let me check. If x was negative, then 4^x would be 1/(4^|x|). But 8 is positive, so that's possible, but 4^x = 8 would require 4^x to be greater than 1, so x must be positive. So negative exponents are not going to work here.

What about fractional exponents? Well, we already considered that with 3/2, which is a fraction. So that's the answer.

Alternatively, maybe using common logarithms? Let me try that. log(4^x) = log(8). Then x log(4) = log(8). Therefore, x = log(8)/log(4). Again, using base 10. Let me compute that.

log(8) is log(2^3) = 3 log(2), and log(4) is log(2^2) = 2 log(2). So again, x = 3 log(2)/2 log(2) = 3/2. Same answer.

So all methods lead to x = 3/2. Therefore, the power of 4 that equals 8 is 3/2.

Let me just verify once more with actual calculation. 4^(3/2). The exponent 3/2 can be thought of as taking the square root first and then cubing, or cubing first and then taking the square root.

Square root of 4 is 2, then cubed is 8. Alternatively, 4 cubed is 64, then square root of 64 is 8. Either way, it's 8. So that's correct.

Therefore, I can be confident that the answer is 3/2.

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

To determine what power of 4 equals 8, we are essentially solving the equation:

$$
4^x = 8
$$

---

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

We know that:

- $ 4 = 2^2 $
- $ 8 = 2^3 $

So, we can rewrite the equation using base 2:

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

---

### Step 2: Apply Exponent Rules

Using the power of a power rule $ (a^b)^c = a^{bc} $, the left-hand side becomes:

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

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

$$
2x = 3
$$

---

### Step 3: Solve for $ x $

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

---

### Step 4: Verify the Solution

Let’s verify that $ 4^{3/2} = 8 $:

$$
4^{3/2} = (4^{1/2})^3 = (\sqrt{4})^3 = 2^3 = 8
$$

This confirms the solution is correct.

---

### Final Answer

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

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. Since the divisor is a linear polynomial, x + 1, synthetic division might be easier. But wait, let me recall how synthetic division works.

First, synthetic division is used when dividing by a linear factor of the form x - c. In this case, the divisor is x + 1, which can be rewritten as x - (-1). So, c would be -1. That seems right. So, I can use synthetic division with c = -1.

But before I jump into synthetic division, maybe I should check if there's another method. Alternatively, I could use polynomial long division. Let me try both methods and see which one works better.

Starting with synthetic division. Let me set up the coefficients of the polynomial x⁶ - 3. The polynomial is x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3. So, the coefficients are 1, 0, 0, 0, 0, 0, -3.

Now, synthetic division steps: write down c = -1, bring down the leading coefficient, multiply by c, add to the next coefficient, and so on. Let me write this out step by step.

Set up:

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

Wait, let me check that again. Let me do each step carefully.

First, bring down the 1.

Multiply 1 by -1, which is -1. Add that to the next coefficient, which is 0: 0 + (-1) = -1.

Multiply -1 by -1, which is 1. Add to the next coefficient, 0 + 1 = 1.

Multiply 1 by -1 = -1. Add to next coefficient, 0 + (-1) = -1.

Multiply -1 by -1 = 1. Add to next coefficient, 0 + 1 = 1.

Multiply 1 by -1 = -1. Add to next coefficient, 0 + (-1) = -1.

Multiply -1 by -1 = 1. Add to the last coefficient, -3 + 1 = -2.

So the result of the synthetic division is the coefficients 1, -1, 1, -1, 1, -1, and the remainder is -2. Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x - 1 with a remainder of -2.

But the question asks for the quotient, so the answer is x⁵ - x⁴ + x³ - x² + x - 1. Let me verify this with another method to make sure I didn't make a mistake.

Alternatively, let's try polynomial long division. Dividing x⁶ - 3 by x + 1.

First, divide x⁶ by x, which gives x⁵. Multiply (x + 1) by x⁵: x⁶ + x⁵. Subtract this from the original polynomial:

(x⁶ - 3) - (x⁶ + x⁵) = -x⁵ - 3.

Next, bring down the next term, but since there are no other terms, we just have -x⁵ - 3. Now divide -x⁵ by x, which is -x⁴. Multiply (x + 1) by -x⁴: -x⁵ - x⁴. Subtract this from the current polynomial:

(-x⁵ - 3) - (-x⁵ - x⁴) = 0 + x⁴ - 3.

Now, bring down the next term, but again, there are no terms. So we have x⁴ - 3. Divide x⁴ by x to get x³. Multiply (x + 1) by x³: x⁴ + x³. Subtract:

(x⁴ - 3) - (x⁴ + x³) = -x³ - 3.

Bring down the next term (which is 0), so we have -x³ + 0x² - 3. Divide -x³ by x to get -x². Multiply (x + 1) by -x²: -x³ - x². Subtract:

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

Bring down the next term (which is 0x), so we have x² + 0x - 3. Divide x² by x to get x. Multiply (x + 1) by x: x² + x. Subtract:

(x² + 0x - 3) - (x² + x) = -x - 3.

Bring down the next term (which is 0), so we have -x + 0x - 3. Divide -x by x to get -1. Multiply (x + 1) by -1: -x -1. Subtract:

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

So the remainder is -2, and the quotient is x⁵ - x⁴ + x³ - x² + x - 1. That matches the result from synthetic division. So that seems correct.

Alternatively, maybe I can use the Remainder Theorem to check the remainder? The Remainder Theorem says that the remainder of dividing by x - c is f(c). Here, we are dividing by x + 1, which is x - (-1), so c = -1. Let me compute f(-1) where f(x) = x⁶ - 3.

f(-1) = (-1)^6 - 3 = 1 - 3 = -2. Which matches the remainder we found earlier. So that's a good check.

Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1.

But let me also think if there's another way to approach this problem. For example, using the fact that x⁶ - 1 can be factored as (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + 2? Wait, no. Wait, x⁶ - 1 is (x³)^2 - 1^2 = (x³ - 1)(x³ + 1) = (x - 1)(x² + x + 1)(x + 1)(x² - x + 1). But maybe that's complicating things.

Alternatively, since we have x⁶ - 3, which is x⁶ - 1 - 2. So, x⁶ - 1 is divisible by x + 1, as we saw, and then subtract 2. Therefore, x⁶ - 3 = (x + 1)(quotient) - 2. Which is consistent with our previous result where the remainder is -2. Therefore, the quotient is the same as the quotient when x⁶ - 1 is divided by x + 1, which is x⁵ - x⁴ + x³ - x² + x - 1, and then subtracting 2 from the original polynomial.

But since we are only asked for the quotient, not the remainder, that's already accounted for.

Alternatively, maybe I can use substitution. Let me think. Let me suppose that x⁶ - 3 = (x + 1)Q(x) + R, where Q(x) is the quotient and R is the remainder. Since we are dividing by a first-degree polynomial, the remainder is a constant. We already found R = -2. So, if I can write Q(x) as (x⁶ - 3 - R)/(x + 1) = (x⁶ - 3 + 2)/(x + 1) = (x⁶ - 1)/(x + 1).

But x⁶ - 1 is a difference of squares: (x³)^2 - 1^2 = (x³ - 1)(x³ + 1). Then, each of those can be factored further.

x³ - 1 = (x - 1)(x² + x + 1), and x³ + 1 = (x + 1)(x² - x + 1). Therefore, x⁶ - 1 = (x - 1)(x² + x + 1)(x + 1)(x² - x + 1). Therefore, (x⁶ - 1)/(x + 1) = (x - 1)(x² + x + 1)(x² - x + 1).

But expanding that would give the quotient Q(x). Let me check if that's equal to x⁵ - x⁴ + x³ - x² + x - 1.

Alternatively, maybe multiply (x² - x + 1)(x² + x + 1) first. Let me compute that.

(x² - x + 1)(x² + x + 1) = [(x² + 1) - x][(x² + 1) + x] = (x² + 1)^2 - x² = x⁴ + 2x² + 1 - x² = x⁴ + x² + 1.

So, then (x⁶ - 1)/(x + 1) = (x - 1)(x⁴ + x² + 1). Let me multiply (x - 1)(x⁴ + x² + 1).

Multiply x*(x⁴ + x² + 1) = x⁵ + x³ + x.

Multiply -1*(x⁴ + x² + 1) = -x⁴ - x² - 1.

Adding those together: x⁵ + x³ + x - x⁴ - x² - 1 = x⁵ - x⁴ + x³ - x² + x - 1.

Which is exactly the quotient we found earlier! So that confirms it. Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1.

Therefore, regardless of the method used—synthetic division, polynomial long division, or factoring—the quotient is the same.

Just to be thorough, let me check with another value. Let me pick a value for x and see if the division works. Let's choose x = 0.

Original polynomial: 0⁶ - 3 = -3. Divided by 0 + 1 = 1. So, -3 divided by 1 is -3.

Now, the quotient when x = 0 is Q(0) = 0⁵ - 0⁴ + 0³ - 0² + 0 - 1 = -1. Then, the remainder is -2. So, according to the division algorithm, f(x) = (x + 1)Q(x) + R. So, plugging x = 0: -3 = (0 + 1)(-1) + (-2) => -3 = -1 - 2 => -3 = -3. Correct.

Another test: x = 1. Original polynomial: 1⁶ - 3 = 1 - 3 = -2. Divided by 1 + 1 = 2. So, -2 / 2 = -1.

Quotient Q(1) = 1 - 1 + 1 - 1 + 1 - 1 = 0. Remainder is -2. So, f(1) = (1 + 1)*Q(1) + R => -2 = 2*0 + (-2) => -2 = -2. Correct.

Another test: x = 2. Original polynomial: 64 - 3 = 61. Divided by 2 + 1 = 3. 61 / 3 is approximately 20.333...

Quotient Q(2) = 32 - 16 + 8 - 4 + 2 - 1 = 32 -16 is 16, +8 is 24, -4 is 20, +2 is 22, -1 is 21. Then remainder is -2. So, according to division algorithm: f(2) = (2 + 1)*Q(2) + R => 61 = 3*21 + (-2) => 63 - 2 = 61. Correct.

So, all these checks work. Therefore, I can be confident that the quotient is correct.

Therefore, the answer is x⁵ - x⁴ + x³ - x² + 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 can use **synthetic division**, which is a streamlined method for dividing a polynomial by a linear factor of the form $ x - c $.

---

### Step 1: Identify the divisor and rewrite it
The divisor is $ x + 1 $, which can be rewritten as $ x - (-1) $. So, we use $ c = -1 $ in synthetic division.

---

### Step 2: Write the coefficients of the dividend
The polynomial $ x^6 - 3 $ has the following coefficients:

$$
1 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad -3
$$

These correspond to the terms $ x^6, x^5, x^4, x^3, x^2, x, $ and the constant term $ -3 $.

---

### Step 3: Perform synthetic division

We set up the synthetic division with $ c = -1 $:

$$
\begin{array}{r|rrrrrrr}
-1 & 1 & 0 & 0 & 0 & 0 & 0 & -3 \\
   &   & -1 & 1 & -1 & 1 & -1 & 1 \\
\hline
   & 1 & -1 & 1 & -1 & 1 & -1 & -2 \\
\end{array}
$$

- Bring down the first coefficient: $ 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to last coefficient: $ -3 + 1 = -2 $

---

### Step 4: Interpret the result

The final row of the synthetic division gives the coefficients of the quotient polynomial, and the last number is the remainder.

So, the quotient is:

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

And the remainder is $ -2 $, which confirms the division 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. Since the divisor is a linear polynomial, x + 1, synthetic division might be easier. But wait, let me recall how synthetic division works.

First, synthetic division is used when dividing by a linear factor of the form x - c. In this case, the divisor is x + 1, which can be rewritten as x - (-1). So, c would be -1. That seems right. So, I can use synthetic division with c = -1.

But before I jump into synthetic division, maybe I should check if there's another method. Alternatively, I could use polynomial long division. Let me try both methods and see which one works better.

Starting with synthetic division. Let me set up the coefficients of the polynomial x⁶ - 3. The polynomial is x⁶ + 0x⁵ + 0x⁴ + 0x³ + 0x² + 0x - 3. So, the coefficients are 1, 0, 0, 0, 0, 0, -3.

Now, synthetic division steps: write down c = -1, bring down the leading coefficient, multiply by c, add to the next coefficient, and so on. Let me write this out step by step.

Set up:

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

Wait, let me check that again. Let me do each step carefully.

First, bring down the 1.

Multiply 1 by -1, which is -1. Add that to the next coefficient, which is 0: 0 + (-1) = -1.

Multiply -1 by -1, which is 1. Add to the next coefficient, 0 + 1 = 1.

Multiply 1 by -1 = -1. Add to next coefficient, 0 + (-1) = -1.

Multiply -1 by -1 = 1. Add to next coefficient, 0 + 1 = 1.

Multiply 1 by -1 = -1. Add to next coefficient, 0 + (-1) = -1.

Multiply -1 by -1 = 1. Add to the last coefficient, -3 + 1 = -2.

So the result of the synthetic division is the coefficients 1, -1, 1, -1, 1, -1, and the remainder is -2. Therefore, the quotient is x⁵ - x⁴ + x³ - x² + x - 1 with a remainder of -2.

But the question asks for the quotient, so the answer is x⁵ - x⁴ + x³ - x² + x - 1. Let me verify this with another method to make sure I didn't make a mistake.

Alternatively, let's try polynomial long division. Dividing x⁶ - 3 by x + 1.

First, divide x⁶ by x, which gives x⁵. Multiply (x + 1) by x⁵: x⁶ + x⁵. Subtract this from the original polynomial:

(x⁶ - 3) - (x⁶ + x⁵) = -x⁵ - 3.

Next, bring down the next term, but since there are no other terms, we just have -x⁵ - 3. Now divide -x⁵ by x, which is -x⁴. Multiply (x + 1) by -x⁴: -x⁵ - x⁴. Subtract this from the current polynomial:

(-x⁵ - 3) - (-x⁵ - x⁴) = 0 + x⁴ - 3.

Now, bring down the next term, but again, there are no terms. So we have x⁴ - 3. Divide x⁴ by x to get x³. Multiply (x + 1) by x³: x⁴ + x³. Subtract:

(x⁴ - 3) - (x⁴ + x³) = -x³ - 3.

Bring down the next term (which is 0), so we have -x³ + 0x² - 3. Divide -x³ by x to get -x². Multiply (x + 1) by -x²: -x³ - x². Subtract:

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

Bring down the next term (which is 0x), so we have x² + 0x - 3. Divide x² by x to get x. Multiply (x + 1) by x: x² + x. Subtract:

(x² + 0x - 3) - (x² + x) = -x - 3.

Bring down the next term (which is 0), so we have -x + 0x - 3. Divide -x by x to get -1. Multiply (x + 1) by -1: -x -1. Subtract:

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

So the remainder is -2, and the quotient is x⁵ - x⁴ + x³ - x² + x - 1. That matches the result from synthetic division. So that seems correct.

Alternatively, maybe I can use the Remainder Theorem to check the remainder? The Remainder Theorem says that the remainder of dividing by x - c is f(c). Here, we are dividing by x + 1, which is x - (-1), so c = -1. Let me compute f(-1) where f(x) = x⁶ - 3.

f(-1) = (-1)^6 - 3 = 1 - 3 = -2. Which matches the remainder we found earlier. So that's a good check.

Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1.

But let me also think if there's another way to approach this problem. For example, using the fact that x⁶ - 1 can be factored as (x + 1)(x⁵ - x⁴ + x³ - x² + x - 1) + 2? Wait, no. Wait, x⁶ - 1 is (x³)^2 - 1^2 = (x³ - 1)(x³ + 1) = (x - 1)(x² + x + 1)(x + 1)(x² - x + 1). But maybe that's complicating things.

Alternatively, since we have x⁶ - 3, which is x⁶ - 1 - 2. So, x⁶ - 1 is divisible by x + 1, as we saw, and then subtract 2. Therefore, x⁶ - 3 = (x + 1)(quotient) - 2. Which is consistent with our previous result where the remainder is -2. Therefore, the quotient is the same as the quotient when x⁶ - 1 is divided by x + 1, which is x⁵ - x⁴ + x³ - x² + x - 1, and then subtracting 2 from the original polynomial.

But since we are only asked for the quotient, not the remainder, that's already accounted for.

Alternatively, maybe I can use substitution. Let me think. Let me suppose that x⁶ - 3 = (x + 1)Q(x) + R, where Q(x) is the quotient and R is the remainder. Since we are dividing by a first-degree polynomial, the remainder is a constant. We already found R = -2. So, if I can write Q(x) as (x⁶ - 3 - R)/(x + 1) = (x⁶ - 3 + 2)/(x + 1) = (x⁶ - 1)/(x + 1).

But x⁶ - 1 is a difference of squares: (x³)^2 - 1^2 = (x³ - 1)(x³ + 1). Then, each of those can be factored further.

x³ - 1 = (x - 1)(x² + x + 1), and x³ + 1 = (x + 1)(x² - x + 1). Therefore, x⁶ - 1 = (x - 1)(x² + x + 1)(x + 1)(x² - x + 1). Therefore, (x⁶ - 1)/(x + 1) = (x - 1)(x² + x + 1)(x² - x + 1).

But expanding that would give the quotient Q(x). Let me check if that's equal to x⁵ - x⁴ + x³ - x² + x - 1.

Alternatively, maybe multiply (x² - x + 1)(x² + x + 1) first. Let me compute that.

(x² - x + 1)(x² + x + 1) = [(x² + 1) - x][(x² + 1) + x] = (x² + 1)^2 - x² = x⁴ + 2x² + 1 - x² = x⁴ + x² + 1.

So, then (x⁶ - 1)/(x + 1) = (x - 1)(x⁴ + x² + 1). Let me multiply (x - 1)(x⁴ + x² + 1).

Multiply x*(x⁴ + x² + 1) = x⁵ + x³ + x.

Multiply -1*(x⁴ + x² + 1) = -x⁴ - x² - 1.

Adding those together: x⁵ + x³ + x - x⁴ - x² - 1 = x⁵ - x⁴ + x³ - x² + x - 1.

Which is exactly the quotient we found earlier! So that confirms it. Therefore, the quotient is indeed x⁵ - x⁴ + x³ - x² + x - 1.

Therefore, regardless of the method used—synthetic division, polynomial long division, or factoring—the quotient is the same.

Just to be thorough, let me check with another value. Let me pick a value for x and see if the division works. Let's choose x = 0.

Original polynomial: 0⁶ - 3 = -3. Divided by 0 + 1 = 1. So, -3 divided by 1 is -3.

Now, the quotient when x = 0 is Q(0) = 0⁵ - 0⁴ + 0³ - 0² + 0 - 1 = -1. Then, the remainder is -2. So, according to the division algorithm, f(x) = (x + 1)Q(x) + R. So, plugging x = 0: -3 = (0 + 1)(-1) + (-2) => -3 = -1 - 2 => -3 = -3. Correct.

Another test: x = 1. Original polynomial: 1⁶ - 3 = 1 - 3 = -2. Divided by 1 + 1 = 2. So, -2 / 2 = -1.

Quotient Q(1) = 1 - 1 + 1 - 1 + 1 - 1 = 0. Remainder is -2. So, f(1) = (1 + 1)*Q(1) + R => -2 = 2*0 + (-2) => -2 = -2. Correct.

Another test: x = 2. Original polynomial: 64 - 3 = 61. Divided by 2 + 1 = 3. 61 / 3 is approximately 20.333...

Quotient Q(2) = 32 - 16 + 8 - 4 + 2 - 1 = 32 -16 is 16, +8 is 24, -4 is 20, +2 is 22, -1 is 21. Then remainder is -2. So, according to division algorithm: f(2) = (2 + 1)*Q(2) + R => 61 = 3*21 + (-2) => 63 - 2 = 61. Correct.

So, all these checks work. Therefore, I can be confident that the quotient is correct.

Therefore, the answer is x⁵ - x⁴ + x³ - x² + 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 can use **synthetic division**, which is a streamlined method for dividing a polynomial by a linear factor of the form $ x - c $.

---

### Step 1: Identify the divisor and rewrite it
The divisor is $ x + 1 $, which can be rewritten as $ x - (-1) $. So, we use $ c = -1 $ in synthetic division.

---

### Step 2: Write the coefficients of the dividend
The polynomial $ x^6 - 3 $ has the following coefficients:

$$
1 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad -3
$$

These correspond to the terms $ x^6, x^5, x^4, x^3, x^2, x, $ and the constant term $ -3 $.

---

### Step 3: Perform synthetic division

We set up the synthetic division with $ c = -1 $:

$$
\begin{array}{r|rrrrrrr}
-1 & 1 & 0 & 0 & 0 & 0 & 0 & -3 \\
   &   & -1 & 1 & -1 & 1 & -1 & 1 \\
\hline
   & 1 & -1 & 1 & -1 & 1 & -1 & -2 \\
\end{array}
$$

- Bring down the first coefficient: $ 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to next coefficient: $ 0 + 1 = 1 $
- Multiply $ 1 \times -1 = -1 $, add to next coefficient: $ 0 + (-1) = -1 $
- Multiply $ -1 \times -1 = 1 $, add to last coefficient: $ -3 + 1 = -2 $

---

### Step 4: Interpret the result

The final row of the synthetic division gives the coefficients of the quotient polynomial, and the last number is the remainder.

So, the quotient is:

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

And the remainder is $ -2 $, which confirms the division 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 the arcsin of -1/2. Hmm, let me recall what arcsin means. Arcsin is the inverse function of sine, right? So, if I have arcsin(x), it means the angle whose sine is x. But I have to remember the domain and range of arcsin to make sure I get the correct answer.

First, the domain of arcsin is between -1 and 1, which is fine because -1/2 is within that range. The range of arcsin is from -π/2 to π/2 radians. That's important because the inverse sine function will only give me angles in that interval. So, I need to find an angle θ in [-π/2, π/2] such that sin(θ) = -1/2.

Let me think about the unit circle. The sine of an angle is -1/2. Sine corresponds to the y-coordinate on the unit circle. So, I need an angle where the y-coordinate is -1/2. I know that sin(π/6) is 1/2, which is positive. But since we need a negative value, the angle must be in a quadrant where sine is negative. The sine function is negative in the third and fourth quadrants. However, the range of arcsin is restricted to -π/2 to π/2, which is the fourth and first quadrants. Wait, actually, -π/2 is pointing downward along the negative y-axis, and π/2 is upward along the positive y-axis. So, angles between -π/2 and 0 are in the fourth quadrant, and between 0 and π/2 are in the first quadrant. Therefore, the angle we are looking for must be in the fourth quadrant because sine is negative there.

So, if sin(π/6) = 1/2, then sin(-π/6) should be -1/2, right? Because sine is an odd function, so sin(-θ) = -sin(θ). Let me check that. If θ is π/6, then sin(-π/6) = -sin(π/6) = -1/2. Yes, that works. And -π/6 is within the range of arcsin, which is between -π/2 and π/2. So, that seems like the answer.

But let me make sure there isn't another angle in the range of arcsin that also satisfies sin(θ) = -1/2. For example, in the third quadrant, the angle would be π + π/6 = 7π/6, but that's outside the range of arcsin because 7π/6 is greater than π/2. Similarly, if I consider angles in the fourth quadrant, the reference angle is π/6, so the angle would be -π/6 or 2π - π/6 = 11π/6. But 11π/6 is more than π/2, so it's not in the range. Therefore, the only angle in the range of arcsin that satisfies sin(θ) = -1/2 is -π/6.

Wait, but let me confirm with the unit circle. If I imagine the unit circle, the angle -π/6 is the same as rotating clockwise π/6 radians from the positive x-axis, which lands me in the fourth quadrant. The coordinates there would be (cos(-π/6), sin(-π/6)) which is (cos(π/6), -sin(π/6)) = (√3/2, -1/2). So, yes, the y-coordinate is -1/2, which matches the sine value. Therefore, that's correct.

Alternatively, if I think about the principal value of arcsin, which is between -π/2 and π/2, so negative angles are allowed. So, -π/6 is indeed in that interval. Therefore, arcsin(-1/2) is -π/6.

But just to be thorough, let me recall some key values. For example, sin(π/6) = 1/2, sin(π/3) = √3/2, sin(π/2) = 1, sin(π) = 0, sin(3π/2) = -1, sin(2π) = 0. So, if I need a sine of -1/2, the reference angle is π/6, and since sine is negative, the angle is in the fourth quadrant, which is -π/6 or equivalently 11π/6, but 11π/6 is more than π/2, so not in the range. Therefore, the answer is -π/6.

Another way to think about it is using the inverse function. If I have a calculator, I could type in arcsin(-1/2), but since I need to do this manually, I have to rely on my knowledge of the unit circle. Let me also remember that arcsin(-x) = -arcsin(x). So, arcsin(-1/2) = -arcsin(1/2). And arcsin(1/2) is π/6, so the answer is -π/6. That's another way to see it, using the odd function property.

Therefore, combining all these thoughts, I can confidently say that arcsin(-1/2) is -π/6 radians.

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

To compute $\arcsin \left( -\frac{1}{2} \right)$, we begin by understanding the definition and properties of the inverse sine function.

---

### Step 1: Understand the Domain and Range of $\arcsin$

The function $\arcsin(x)$ is defined for $x \in [-1, 1]$, and its **range** is restricted to $[- \frac{\pi}{2}, \frac{\pi}{2}]$ radians. This means that the output of $\arcsin(x)$ is always an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive.

---

### Step 2: Use the Odd Function Property

Recall that the sine function is **odd**, meaning:

$$
\sin(-\theta) = -\sin(\theta)
$$

This implies:

$$
\arcsin(-x) = -\arcsin(x)
$$

So, applying this property:

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

---

### Step 3: Evaluate $\arcsin\left(\frac{1}{2}\right)$

From standard trigonometric values, we know:

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

Therefore:

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

---

### Step 4: Apply the Negative Sign

Now, using the earlier result:

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

---

### Step 5: Verify the Result Lies in the Range

The value $-\frac{\pi}{6}$ is within the range of $\arcsin$, which is $[- \frac{\pi}{2}, \frac{\pi}{2}]$, so it is a valid output.

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to compute the arcsin of -1/2. Hmm, let me recall what arcsin means. Arcsin is the inverse function of sine, right? So, if I have arcsin(x), it means the angle whose sine is x. But I have to remember the domain and range of arcsin to make sure I get the correct answer.

First, the domain of arcsin is between -1 and 1, which is fine because -1/2 is within that range. The range of arcsin is from -π/2 to π/2 radians. That's important because the inverse sine function will only give me angles in that interval. So, I need to find an angle θ in [-π/2, π/2] such that sin(θ) = -1/2.

Let me think about the unit circle. The sine of an angle is -1/2. Sine corresponds to the y-coordinate on the unit circle. So, I need an angle where the y-coordinate is -1/2. I know that sin(π/6) is 1/2, which is positive. But since we need a negative value, the angle must be in a quadrant where sine is negative. The sine function is negative in the third and fourth quadrants. However, the range of arcsin is restricted to -π/2 to π/2, which is the fourth and first quadrants. Wait, actually, -π/2 is pointing downward along the negative y-axis, and π/2 is upward along the positive y-axis. So, angles between -π/2 and 0 are in the fourth quadrant, and between 0 and π/2 are in the first quadrant. Therefore, the angle we are looking for must be in the fourth quadrant because sine is negative there.

So, if sin(π/6) = 1/2, then sin(-π/6) should be -1/2, right? Because sine is an odd function, so sin(-θ) = -sin(θ). Let me check that. If θ is π/6, then sin(-π/6) = -sin(π/6) = -1/2. Yes, that works. And -π/6 is within the range of arcsin, which is between -π/2 and π/2. So, that seems like the answer.

But let me make sure there isn't another angle in the range of arcsin that also satisfies sin(θ) = -1/2. For example, in the third quadrant, the angle would be π + π/6 = 7π/6, but that's outside the range of arcsin because 7π/6 is greater than π/2. Similarly, if I consider angles in the fourth quadrant, the reference angle is π/6, so the angle would be -π/6 or 2π - π/6 = 11π/6. But 11π/6 is more than π/2, so it's not in the range. Therefore, the only angle in the range of arcsin that satisfies sin(θ) = -1/2 is -π/6.

Wait, but let me confirm with the unit circle. If I imagine the unit circle, the angle -π/6 is the same as rotating clockwise π/6 radians from the positive x-axis, which lands me in the fourth quadrant. The coordinates there would be (cos(-π/6), sin(-π/6)) which is (cos(π/6), -sin(π/6)) = (√3/2, -1/2). So, yes, the y-coordinate is -1/2, which matches the sine value. Therefore, that's correct.

Alternatively, if I think about the principal value of arcsin, which is between -π/2 and π/2, so negative angles are allowed. So, -π/6 is indeed in that interval. Therefore, arcsin(-1/2) is -π/6.

But just to be thorough, let me recall some key values. For example, sin(π/6) = 1/2, sin(π/3) = √3/2, sin(π/2) = 1, sin(π) = 0, sin(3π/2) = -1, sin(2π) = 0. So, if I need a sine of -1/2, the reference angle is π/6, and since sine is negative, the angle is in the fourth quadrant, which is -π/6 or equivalently 11π/6, but 11π/6 is more than π/2, so not in the range. Therefore, the answer is -π/6.

Another way to think about it is using the inverse function. If I have a calculator, I could type in arcsin(-1/2), but since I need to do this manually, I have to rely on my knowledge of the unit circle. Let me also remember that arcsin(-x) = -arcsin(x). So, arcsin(-1/2) = -arcsin(1/2). And arcsin(1/2) is π/6, so the answer is -π/6. That's another way to see it, using the odd function property.

Therefore, combining all these thoughts, I can confidently say that arcsin(-1/2) is -π/6 radians.

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

To compute $\arcsin \left( -\frac{1}{2} \right)$, we begin by understanding the definition and properties of the inverse sine function.

---

### Step 1: Understand the Domain and Range of $\arcsin$

The function $\arcsin(x)$ is defined for $x \in [-1, 1]$, and its **range** is restricted to $[- \frac{\pi}{2}, \frac{\pi}{2}]$ radians. This means that the output of $\arcsin(x)$ is always an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, inclusive.

---

### Step 2: Use the Odd Function Property

Recall that the sine function is **odd**, meaning:

$$
\sin(-\theta) = -\sin(\theta)
$$

This implies:

$$
\arcsin(-x) = -\arcsin(x)
$$

So, applying this property:

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

---

### Step 3: Evaluate $\arcsin\left(\frac{1}{2}\right)$

From standard trigonometric values, we know:

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

Therefore:

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

---

### Step 4: Apply the Negative Sign

Now, using the earlier result:

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

---

### Step 5: Verify the Result Lies in the Range

The value $-\frac{\pi}{6}$ is within the range of $\arcsin$, which is $[- \frac{\pi}{2}, \frac{\pi}{2}]$, so it is a valid output.

---

### 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 a fraction to a decimal involves dividing the numerator by the denominator. So in this case, I need to divide 3 by 20.

Wait, let me make sure I remember that correctly. Yeah, fractions are just another way of writing division, right? So 3/20 is the same as 3 divided by 20. Alright, so I need to perform that division. But how exactly do I do that? Let me recall.

I think there are a couple of ways to do this. One way is to use long division, where I divide 3 by 20. But since 20 is larger than 3, the result will be less than 1. Maybe I can add a decimal point and some zeros to the numerator to make the division possible. Let me try that.

So, if I write 3 divided by 20, I can think of it as 3.000 divided by 20. Let me set that up.

First, 20 goes into 3 zero times. So I put a 0. and then proceed. Then, 20 goes into 30 once, right? Because 20 times 1 is 20. Subtract 20 from 30, and I get 10. Then bring down the next 0, making it 100.

Wait, hold on. Let me write this step by step.

Starting with 3 divided by 20:

- 20 | 3.000

Since 20 doesn't go into 3, we write 0. and then consider 30 (by adding a decimal point and a zero).

So, 20 goes into 30 once (1). Multiply 1 by 20, subtract from 30, we get 10. Bring down the next 0, making it 100.

Now, 20 goes into 100 five times (since 20*5=100). Subtract 100 from 100, we get 0. So there's no remainder.

Therefore, putting it all together, we have 0.15.

Wait, so 3 divided by 20 is 0.15? Let me check that.

Alternatively, maybe I can think of 3/20 as a fraction that can be converted to a decimal by multiplying numerator and denominator by something to make the denominator a power of 10.

Since 20 is a factor of 100, maybe I can multiply numerator and denominator by 5. Let me try that.

Multiplying numerator and denominator by 5:

3 * 5 = 15

20 * 5 = 100

So, 3/20 = 15/100.

And 15/100 is 0.15.

Oh, that's a quicker way. So that confirms the previous result.

Alternatively, I can think of 3/20 as 3 divided by 2 divided by 10.

Since dividing by 20 is the same as dividing by 2 and then dividing by 10.

So, 3 divided by 2 is 1.5, and then dividing by 10 moves the decimal one place to the left, giving 0.15.

That's another way to see it.

So all three methods give me 0.15.

Let me verify with a calculator to be sure.

Well, if I take 3 divided by 20, 20 times 0.1 is 2, 20 times 0.15 is 3. So yes, 0.15 times 20 is 3.

Therefore, 3/20 as a decimal is 0.15.

I think that's solid. All different methods lead to the same answer, so I'm confident that's correct.

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

To convert the fraction $\frac{3}{20}$ into a decimal, we can use the fundamental concept that a fraction represents division. That is, $\frac{3}{20}$ is equivalent to $3 \div 20$.

---

### Step-by-Step Division

We perform the division $3 \div 20$:

1. Since 20 is greater than 3, the result will be less than 1. We add a decimal point and zeros to the numerator to continue the division.
2. $20$ goes into $30$ once (since $20 \times 1 = 20$).
3. Subtract $20$ from $30$, we get a remainder of $10$.
4. Bring down the next zero, making it $100$.
5. $20$ goes into $100$ exactly five times (since $20 \times 5 = 100$).
6. Subtract $100$ from $100$, and the remainder is $0$.

So, the result of the division is $0.15$.

---

### Alternative Method: Converting to a Decimal by Adjusting the Denominator

Another way to convert $\frac{3}{20}$ to a decimal is to adjust the denominator to a power of 10. Since $20 \times 5 = 100$, we multiply both numerator and denominator by 5:

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

Now, $\frac{15}{100}$ is simply $0.15$.

---

### Final Answer

$$
\boxed{0.15}
$$

[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 a fraction to a decimal involves dividing the numerator by the denominator. So in this case, I need to divide 3 by 20.

Wait, let me make sure I remember that correctly. Yeah, fractions are just another way of writing division, right? So 3/20 is the same as 3 divided by 20. Alright, so I need to perform that division. But how exactly do I do that? Let me recall.

I think there are a couple of ways to do this. One way is to use long division, where I divide 3 by 20. But since 20 is larger than 3, the result will be less than 1. Maybe I can add a decimal point and some zeros to the numerator to make the division possible. Let me try that.

So, if I write 3 divided by 20, I can think of it as 3.000 divided by 20. Let me set that up.

First, 20 goes into 3 zero times. So I put a 0. and then proceed. Then, 20 goes into 30 once, right? Because 20 times 1 is 20. Subtract 20 from 30, and I get 10. Then bring down the next 0, making it 100.

Wait, hold on. Let me write this step by step.

Starting with 3 divided by 20:

- 20 | 3.000

Since 20 doesn't go into 3, we write 0. and then consider 30 (by adding a decimal point and a zero).

So, 20 goes into 30 once (1). Multiply 1 by 20, subtract from 30, we get 10. Bring down the next 0, making it 100.

Now, 20 goes into 100 five times (since 20*5=100). Subtract 100 from 100, we get 0. So there's no remainder.

Therefore, putting it all together, we have 0.15.

Wait, so 3 divided by 20 is 0.15? Let me check that.

Alternatively, maybe I can think of 3/20 as a fraction that can be converted to a decimal by multiplying numerator and denominator by something to make the denominator a power of 10.

Since 20 is a factor of 100, maybe I can multiply numerator and denominator by 5. Let me try that.

Multiplying numerator and denominator by 5:

3 * 5 = 15

20 * 5 = 100

So, 3/20 = 15/100.

And 15/100 is 0.15.

Oh, that's a quicker way. So that confirms the previous result.

Alternatively, I can think of 3/20 as 3 divided by 2 divided by 10.

Since dividing by 20 is the same as dividing by 2 and then dividing by 10.

So, 3 divided by 2 is 1.5, and then dividing by 10 moves the decimal one place to the left, giving 0.15.

That's another way to see it.

So all three methods give me 0.15.

Let me verify with a calculator to be sure.

Well, if I take 3 divided by 20, 20 times 0.1 is 2, 20 times 0.15 is 3. So yes, 0.15 times 20 is 3.

Therefore, 3/20 as a decimal is 0.15.

I think that's solid. All different methods lead to the same answer, so I'm confident that's correct.

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

To convert the fraction $\frac{3}{20}$ into a decimal, we can use the fundamental concept that a fraction represents division. That is, $\frac{3}{20}$ is equivalent to $3 \div 20$.

---

### Step-by-Step Division

We perform the division $3 \div 20$:

1. Since 20 is greater than 3, the result will be less than 1. We add a decimal point and zeros to the numerator to continue the division.
2. $20$ goes into $30$ once (since $20 \times 1 = 20$).
3. Subtract $20$ from $30$, we get a remainder of $10$.
4. Bring down the next zero, making it $100$.
5. $20$ goes into $100$ exactly five times (since $20 \times 5 = 100$).
6. Subtract $100$ from $100$, and the remainder is $0$.

So, the result of the division is $0.15$.

---

### Alternative Method: Converting to a Decimal by Adjusting the Denominator

Another way to convert $\frac{3}{20}$ to a decimal is to adjust the denominator to a power of 10. Since $20 \times 5 = 100$, we multiply both numerator and denominator by 5:

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

Now, $\frac{15}{100}$ is simply $0.15$.

---

### Final Answer

$$
\boxed{0.15}
$$

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? Or maybe there's another way? Wait, I also recall that when you raise a complex number to a power, you can use De Moivre's theorem, which involves the modulus and the argument. Let me try both methods and see if they match.

First, let me compute the modulus of (1 - i). The modulus is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So, |1 - i| = sqrt(2). Then, if I raise this modulus to the 8th power, it would be (sqrt(2))⁸. Let me compute that. Since sqrt(2) is 2^(1/2), raising it to the 8th power would be 2^(1/2 * 8) = 2^4 = 16. So, that would be 16. Is that right? Let me check with another method.

Alternatively, using De Moivre's theorem. First, express 1 - i in polar form. The modulus is sqrt(2) as before. The argument θ is the angle made with the positive real axis. Since the complex number is 1 - i, it's in the fourth quadrant. The reference angle would be arctangent of |b/a| = arctan(1/1) = π/4. So, the argument θ is -π/4 or 7π/4. So, in polar form, 1 - i is sqrt(2)*(cos(-π/4) + i sin(-π/4)).

Now, raising this to the 8th power: [sqrt(2)]^8 * [cos(8*(-π/4)) + i sin(8*(-π/4))]. Wait, but the modulus part is [sqrt(2)]^8 which we already calculated as 16. Then the angle part is 8*(-π/4) = -2π. But cosine and sine are periodic with period 2π, so cos(-2π) = cos(0) = 1, and sin(-2π) = sin(0) = 0. Therefore, the result is 16*(1 + 0i) = 16. So that matches the previous result. So, the modulus is 16. Wait, but the question is asking for |(1 - i)^8|. So, is that the same as |1 - i|^8? Yes, because modulus of a power is the power of the modulus. So, |z^n| = |z|^n. Therefore, |(1 - i)^8| = |1 - i|^8 = (sqrt(2))^8 = 16. So, that seems straightforward.

But let me verify by expanding (1 - i)^8 directly, just to be thorough. Although that might take longer, but maybe I can do it step by step. Let me recall that (a - b)^2 = a² - 2ab + b². So, (1 - i)^2 = 1² - 2*1*i + i² = 1 - 2i + (-1) = 0 - 2i = -2i. Then, (1 - i)^4 = [(1 - i)^2]^2 = (-2i)^2 = (-2)^2 * i² = 4 * (-1) = -4. Then, (1 - i)^8 = [(1 - i)^4]^2 = (-4)^2 = 16. Oh, that's even simpler! So, that gives the same answer. So, regardless of the method, it's 16. Therefore, Ringo should have found 16.

Wait, but let me check once more. Maybe I made a mistake in the expansion? Let me recompute (1 - i)^2. 1 squared is 1, minus 2 times 1 times i is -2i, plus i squared is -1. So, 1 - 2i -1 = -2i. Correct. Then (1 - i)^4 is (-2i)^2. Squaring -2i: (-2i)*(-2i) = 4i² = 4*(-1) = -4. Correct. Then (1 - i)^8 is (-4)^2 = 16. Yep, that's right. So, all methods lead to 16. Therefore, the answer is 16.

Alternatively, if I use binomial expansion for (1 - i)^8, but that would be tedious. Let me try a smaller exponent to check. For example, (1 - i)^1 is 1 - i. The modulus is sqrt(2). (1 - i)^2 is -2i, modulus is 2. (1 - i)^3 would be (1 - i)^2*(1 - i) = (-2i)(1 - i) = -2i + 2i² = -2i - 2. The modulus is sqrt((-2)^2 + (-2)^2) = sqrt(4 + 4) = sqrt(8) = 2*sqrt(2). Which is |1 - i|^3 = (sqrt(2))^3 = 2^(3/2) = 2*sqrt(2). Correct. Similarly, (1 - i)^4 modulus is |1 - i|^4 = (sqrt(2))^4 = 4. Which matches our previous result of -4, modulus 4. So, that's consistent. Therefore, (1 - i)^8 modulus is 16.

Therefore, I think the answer is definitely 16. So, Ringo should have found 16.

**Final Answer**
\boxed{16}

To solve the expression $ |(1 - i)^8| $, we can approach the problem using properties of complex numbers and their modulus.

---

### Step 1: Understand the Modulus of a Complex Number

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

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

In this case, the complex number is $ 1 - i $, so:

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

---

### Step 2: Use the Property of Modulus with Powers

A key property of modulus is:

$$
|z^n| = |z|^n
$$

So, applying this to our problem:

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

Now compute $ (\sqrt{2})^8 $:

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

---

### Step 3: Verification via Direct Expansion (Optional)

We can also verify this by expanding $ (1 - i)^8 $ step by step:

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

This confirms the result.

---

### 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? Or maybe there's another way? Wait, I also recall that when you raise a complex number to a power, you can use De Moivre's theorem, which involves the modulus and the argument. Let me try both methods and see if they match.

First, let me compute the modulus of (1 - i). The modulus is sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). So, |1 - i| = sqrt(2). Then, if I raise this modulus to the 8th power, it would be (sqrt(2))⁸. Let me compute that. Since sqrt(2) is 2^(1/2), raising it to the 8th power would be 2^(1/2 * 8) = 2^4 = 16. So, that would be 16. Is that right? Let me check with another method.

Alternatively, using De Moivre's theorem. First, express 1 - i in polar form. The modulus is sqrt(2) as before. The argument θ is the angle made with the positive real axis. Since the complex number is 1 - i, it's in the fourth quadrant. The reference angle would be arctangent of |b/a| = arctan(1/1) = π/4. So, the argument θ is -π/4 or 7π/4. So, in polar form, 1 - i is sqrt(2)*(cos(-π/4) + i sin(-π/4)).

Now, raising this to the 8th power: [sqrt(2)]^8 * [cos(8*(-π/4)) + i sin(8*(-π/4))]. Wait, but the modulus part is [sqrt(2)]^8 which we already calculated as 16. Then the angle part is 8*(-π/4) = -2π. But cosine and sine are periodic with period 2π, so cos(-2π) = cos(0) = 1, and sin(-2π) = sin(0) = 0. Therefore, the result is 16*(1 + 0i) = 16. So that matches the previous result. So, the modulus is 16. Wait, but the question is asking for |(1 - i)^8|. So, is that the same as |1 - i|^8? Yes, because modulus of a power is the power of the modulus. So, |z^n| = |z|^n. Therefore, |(1 - i)^8| = |1 - i|^8 = (sqrt(2))^8 = 16. So, that seems straightforward.

But let me verify by expanding (1 - i)^8 directly, just to be thorough. Although that might take longer, but maybe I can do it step by step. Let me recall that (a - b)^2 = a² - 2ab + b². So, (1 - i)^2 = 1² - 2*1*i + i² = 1 - 2i + (-1) = 0 - 2i = -2i. Then, (1 - i)^4 = [(1 - i)^2]^2 = (-2i)^2 = (-2)^2 * i² = 4 * (-1) = -4. Then, (1 - i)^8 = [(1 - i)^4]^2 = (-4)^2 = 16. Oh, that's even simpler! So, that gives the same answer. So, regardless of the method, it's 16. Therefore, Ringo should have found 16.

Wait, but let me check once more. Maybe I made a mistake in the expansion? Let me recompute (1 - i)^2. 1 squared is 1, minus 2 times 1 times i is -2i, plus i squared is -1. So, 1 - 2i -1 = -2i. Correct. Then (1 - i)^4 is (-2i)^2. Squaring -2i: (-2i)*(-2i) = 4i² = 4*(-1) = -4. Correct. Then (1 - i)^8 is (-4)^2 = 16. Yep, that's right. So, all methods lead to 16. Therefore, the answer is 16.

Alternatively, if I use binomial expansion for (1 - i)^8, but that would be tedious. Let me try a smaller exponent to check. For example, (1 - i)^1 is 1 - i. The modulus is sqrt(2). (1 - i)^2 is -2i, modulus is 2. (1 - i)^3 would be (1 - i)^2*(1 - i) = (-2i)(1 - i) = -2i + 2i² = -2i - 2. The modulus is sqrt((-2)^2 + (-2)^2) = sqrt(4 + 4) = sqrt(8) = 2*sqrt(2). Which is |1 - i|^3 = (sqrt(2))^3 = 2^(3/2) = 2*sqrt(2). Correct. Similarly, (1 - i)^4 modulus is |1 - i|^4 = (sqrt(2))^4 = 4. Which matches our previous result of -4, modulus 4. So, that's consistent. Therefore, (1 - i)^8 modulus is 16.

Therefore, I think the answer is definitely 16. So, Ringo should have found 16.

**Final Answer**
\boxed{16}

To solve the expression $ |(1 - i)^8| $, we can approach the problem using properties of complex numbers and their modulus.

---

### Step 1: Understand the Modulus of a Complex Number

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

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

In this case, the complex number is $ 1 - i $, so:

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

---

### Step 2: Use the Property of Modulus with Powers

A key property of modulus is:

$$
|z^n| = |z|^n
$$

So, applying this to our problem:

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

Now compute $ (\sqrt{2})^8 $:

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

---

### Step 3: Verification via Direct Expansion (Optional)

We can also verify this by expanding $ (1 - i)^8 $ step by step:

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

This confirms the result.

---

### 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.


Explanation:
Okay, so there's this problem about a class of 50 students. Some participate in MATHCOUNTS, some in the science club, and some in neither. The question is asking how many students participate in both. Let me try to figure this out step by step.

First, let me recall some set theory principles. I think this is a classic inclusion-exclusion problem. The formula for two sets is: the number of elements in either set A or set B is equal to the number in A plus the number in B minus the number in both. So, |A ∪ B| = |A| + |B| - |A ∩ B|.

Alright, so applying that here. Let me define the sets. Let me let M be the set of students in MATHCOUNTS, and S be the set of students in the science club. The total number of students is 50. We know that 6 students participate in neither, so that means the number of students who are in at least one of the two groups (MATHCOUNTS or science club) is 50 - 6 = 44.

So, |M ∪ S| = 44.

We also know that |M| is 28 and |S| is 21. So, plugging into the inclusion-exclusion formula:

|M ∪ S| = |M| + |S| - |M ∩ S|

So, 44 = 28 + 21 - |M ∩ S|

Let me compute 28 + 21. That's 49. So:

44 = 49 - |M ∩ S|

Hmm, so if I rearrange that equation, subtract 49 from both sides:

44 - 49 = - |M ∩ S|

Which is -5 = - |M ∩ S|

Multiply both sides by -1:

5 = |M ∩ S|

So, that would mean that 5 students are in both MATHCOUNTS and the science club. Let me check if that makes sense.

If there are 28 in MATHCOUNTS and 5 in both, then the number only in MATHCOUNTS is 28 - 5 = 23. Similarly, the number only in science club is 21 - 5 = 16. Then, adding those who are only in MATHCOUNTS, only in science club, and both, we should get the total number in at least one group. So 23 + 16 + 5 = 44. Which matches the 50 - 6 = 44. So that seems to check out.

Wait, let me verify again. Total students: 50. 6 in neither. So 44 in at least one. If 23 only in MATHCOUNTS, 16 only in science, and 5 in both, then total is 23 + 16 + 5 = 44. Yes, that's correct. So the number in both is 5.

But let me think again if I did the formula correctly. The formula says |M ∪ S| = |M| + |S| - |M ∩ S|. So solving for |M ∩ S|, it's |M| + |S| - |M ∪ S|. Which is 28 + 21 - 44 = 49 - 44 = 5. Yep, that's the same. So that seems correct.

Alternatively, maybe I can think of it as a Venn diagram. There's two circles overlapping. The total in the union is 44. The sum of the individual groups is 28 + 21 = 49. But since the union is only 44, the overlap is 49 - 44 = 5. So that's another way to think about it.

Is there another way to approach this? Maybe using a table or something? Let me try.

Total students: 50

Students in neither: 6

So students in at least one: 50 - 6 = 44

Let me make a table with rows for MATHCOUNTS and Science Club, and columns for Yes and No.

But maybe that's complicating. Alternatively, think of it as:

Number in MATHCOUNTS only: ?

Number in Science Club only: ?

Number in both: ?

Number in neither: 6

Total: 50

So, if I let x be the number in both, then:

Number in MATHCOUNTS only is 28 - x

Number in Science Club only is 21 - x

Then total in at least one is (28 - x) + (21 - x) + x = 28 - x + 21 - x + x = 28 + 21 - x = 49 - x

But we know that total in at least one is 44, so 49 - x = 44 => x = 5. Yep, same answer.

So that seems consistent. So I think 5 is the correct answer.

Wait, but let me check if there's any possibility that the numbers could be different. For example, if there was some overlap that's not considered? But no, the formula is pretty straightforward. The inclusion-exclusion principle is pretty fundamental here.

Alternatively, maybe someone might think that since 28 + 21 = 49, which is 5 more than 44, so the overlap is 5. That's exactly what we did. So that seems to make sense.

Another sanity check: if there were no overlap, the total number in at least one would be 28 + 21 = 49, but since there are only 44, that means there's an overlap of 5. So that's another way to think about it.

So, yeah, I think 5 is the right answer.

**Final Answer**
The number of students who participate in both MATHCOUNTS and science club is \boxed{5}.

To determine how many students participate in both MATHCOUNTS and the science club, we can use the **principle of inclusion and exclusion** from set theory.

---

### Step 1: Understand the Given Information

- Total number of students: **50**
- Students in **MATHCOUNTS**: **28**
- Students in **science club**: **21**
- Students in **neither**: **6**

From this, we can find the number of students who are in **at least one** of the two groups:

$$
\text{Students in at least one group} = 50 - 6 = 44
$$

---

### Step 2: Apply the Inclusion-Exclusion Principle

Let:
- $ |M| = 28 $: number of students in MATHCOUNTS
- $ |S| = 21 $: number of students in science club
- $ |M \cup S| = 44 $: number of students in at least one group
- $ |M \cap S| = x $: number of students in **both** groups (what we're solving for)

Using the inclusion-exclusion formula:

$$
|M \cup S| = |M| + |S| - |M \cap S|
$$

Substitute the known values:

$$
44 = 28 + 21 - x
$$

$$
44 = 49 - x
$$

$$
x = 49 - 44 = 5
$$

---

### Step 3: Interpret the Result

This means that **5 students** are participating in **both** MATHCOUNTS and the science club.

---

### Final Answer

$$
\boxed{5}
$$

[Assistant]
Explanation:
Okay, so there's this problem about a class of 50 students. Some participate in MATHCOUNTS, some in the science club, and some in neither. The question is asking how many students participate in both. Let me try to figure this out step by step.

First, let me recall some set theory principles. I think this is a classic inclusion-exclusion problem. The formula for two sets is: the number of elements in either set A or set B is equal to the number in A plus the number in B minus the number in both. So, |A ∪ B| = |A| + |B| - |A ∩ B|.

Alright, so applying that here. Let me define the sets. Let me let M be the set of students in MATHCOUNTS, and S be the set of students in the science club. The total number of students is 50. We know that 6 students participate in neither, so that means the number of students who are in at least one of the two groups (MATHCOUNTS or science club) is 50 - 6 = 44.

So, |M ∪ S| = 44.

We also know that |M| is 28 and |S| is 21. So, plugging into the inclusion-exclusion formula:

|M ∪ S| = |M| + |S| - |M ∩ S|

So, 44 = 28 + 21 - |M ∩ S|

Let me compute 28 + 21. That's 49. So:

44 = 49 - |M ∩ S|

Hmm, so if I rearrange that equation, subtract 49 from both sides:

44 - 49 = - |M ∩ S|

Which is -5 = - |M ∩ S|

Multiply both sides by -1:

5 = |M ∩ S|

So, that would mean that 5 students are in both MATHCOUNTS and the science club. Let me check if that makes sense.

If there are 28 in MATHCOUNTS and 5 in both, then the number only in MATHCOUNTS is 28 - 5 = 23. Similarly, the number only in science club is 21 - 5 = 16. Then, adding those who are only in MATHCOUNTS, only in science club, and both, we should get the total number in at least one group. So 23 + 16 + 5 = 44. Which matches the 50 - 6 = 44. So that seems to check out.

Wait, let me verify again. Total students: 50. 6 in neither. So 44 in at least one. If 23 only in MATHCOUNTS, 16 only in science, and 5 in both, then total is 23 + 16 + 5 = 44. Yes, that's correct. So the number in both is 5.

But let me think again if I did the formula correctly. The formula says |M ∪ S| = |M| + |S| - |M ∩ S|. So solving for |M ∩ S|, it's |M| + |S| - |M ∪ S|. Which is 28 + 21 - 44 = 49 - 44 = 5. Yep, that's the same. So that seems correct.

Alternatively, maybe I can think of it as a Venn diagram. There's two circles overlapping. The total in the union is 44. The sum of the individual groups is 28 + 21 = 49. But since the union is only 44, the overlap is 49 - 44 = 5. So that's another way to think about it.

Is there another way to approach this? Maybe using a table or something? Let me try.

Total students: 50

Students in neither: 6

So students in at least one: 50 - 6 = 44

Let me make a table with rows for MATHCOUNTS and Science Club, and columns for Yes and No.

But maybe that's complicating. Alternatively, think of it as:

Number in MATHCOUNTS only: ?

Number in Science Club only: ?

Number in both: ?

Number in neither: 6

Total: 50

So, if I let x be the number in both, then:

Number in MATHCOUNTS only is 28 - x

Number in Science Club only is 21 - x

Then total in at least one is (28 - x) + (21 - x) + x = 28 - x + 21 - x + x = 28 + 21 - x = 49 - x

But we know that total in at least one is 44, so 49 - x = 44 => x = 5. Yep, same answer.

So that seems consistent. So I think 5 is the correct answer.

Wait, but let me check if there's any possibility that the numbers could be different. For example, if there was some overlap that's not considered? But no, the formula is pretty straightforward. The inclusion-exclusion principle is pretty fundamental here.

Alternatively, maybe someone might think that since 28 + 21 = 49, which is 5 more than 44, so the overlap is 5. That's exactly what we did. So that seems to make sense.

Another sanity check: if there were no overlap, the total number in at least one would be 28 + 21 = 49, but since there are only 44, that means there's an overlap of 5. So that's another way to think about it.

So, yeah, I think 5 is the right answer.

**Final Answer**
The number of students who participate in both MATHCOUNTS and science club is \boxed{5}.

To determine how many students participate in both MATHCOUNTS and the science club, we can use the **principle of inclusion and exclusion** from set theory.

---

### Step 1: Understand the Given Information

- Total number of students: **50**
- Students in **MATHCOUNTS**: **28**
- Students in **science club**: **21**
- Students in **neither**: **6**

From this, we can find the number of students who are in **at least one** of the two groups:

$$
\text{Students in at least one group} = 50 - 6 = 44
$$

---

### Step 2: Apply the Inclusion-Exclusion Principle

Let:
- $ |M| = 28 $: number of students in MATHCOUNTS
- $ |S| = 21 $: number of students in science club
- $ |M \cup S| = 44 $: number of students in at least one group
- $ |M \cap S| = x $: number of students in **both** groups (what we're solving for)

Using the inclusion-exclusion formula:

$$
|M \cup S| = |M| + |S| - |M \cap S|
$$

Substitute the known values:

$$
44 = 28 + 21 - x
$$

$$
44 = 49 - x
$$

$$
x = 49 - 44 = 5
$$

---

### Step 3: Interpret the Result

This means that **5 students** are participating in **both** MATHCOUNTS and the science club.

---

### Final Answer

$$
\boxed{5}
$$

Solution: Among the $50-6=44$ students that participate in either MATHCOUNTS or science club, $44-28=16$ students do not participate in MATHCOUNTS.  All 16 of these students only participate in science club.  The other $21-16=\boxed{5}$ science club participants participate in MATHCOUNTS as well.