Explanation:
Okay, so I need to find the value of N where the sum of the first N positive odd numbers is 121. Hmm, let me think. I remember that there's a formula for the sum of the first N odd numbers. Wait, isn't it something like N squared? Let me check that.

Let me start by recalling some examples. The first odd number is 1. If N is 1, the sum is 1. According to the formula, 1 squared is 1. That works. The first two odd numbers are 1 and 3. Their sum is 4, which is 2 squared. The first three odd numbers: 1 + 3 + 5 = 9, which is 3 squared. Oh, right! So the pattern seems to be that the sum of the first N odd numbers is N². So if the sum is 121, then N² = 121. Therefore, N should be the square root of 121. Let me calculate that. The square root of 121 is 11, because 11 times 11 is 121. So N is 11? Let me verify this to make sure I didn't make a mistake.

Let me add up the first 11 odd numbers to check. The sequence of odd numbers is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. Let me add them step by step.

Starting with 1 + 3 = 4. Then 4 + 5 = 9. 9 + 7 = 16. 16 + 9 = 25. 25 + 11 = 36. 36 + 13 = 49. 49 + 15 = 64. 64 + 17 = 81. 81 + 19 = 100. 100 + 21 = 121. Wow, that adds up to 121. So N is indeed 11.

But just to make sure I didn't skip any steps or make an arithmetic error, let me try another way. Maybe using the formula for the sum of an arithmetic series. The sum of an arithmetic series is (number of terms)/2 times (first term + last term).

The first term is 1, and the Nth term is 2N - 1. So the sum would be N/2 * (1 + (2N - 1)) = N/2 * (2N) = N * N = N². Yep, that confirms the formula. So if the sum is 121, then N² = 121, so N = sqrt(121) = 11.

Alternatively, maybe I can think about the fact that each time you add an odd number, you're essentially building a square. Like, 1 is 1x1, then adding 3 gives 2x2, adding 5 gives 3x3, and so on. So the Nth odd number is the one that makes the square go from (N-1)² to N². Therefore, the sum of the first N odd numbers is N².

So, if the sum is 121, then N is 11.

Wait, but just to be thorough, let me check if there's any possibility that N could be negative? But since N is the number of terms, it has to be a positive integer. So N = 11 is the only solution.

Another way to think about it: if I didn't remember the formula, I could derive it. Let's say the sum S = 1 + 3 + 5 + ... + (2N - 1). Let me write this out:

S = 1 + 3 + 5 + ... + (2N - 1)

If I add the first term and the last term: 1 + (2N - 1) = 2N

Similarly, the second term and the second last term: 3 + (2N - 3) = 2N

And so on. Since there are N terms, there are N/2 such pairs if N is even, or (N-1)/2 pairs plus the middle term if N is odd. But in this case, since all the pairs add up to 2N, and there are N terms, which is either even or odd. Wait, if N is even, then there are N/2 pairs, each adding to 2N, so total sum is N/2 * 2N = N². If N is odd, then there are (N-1)/2 pairs adding to 2N, and the middle term is N. So total sum is (N-1)/2 * 2N + N = N(N - 1) + N = N² - N + N = N². So regardless of whether N is even or odd, the sum is N². Therefore, the formula holds.

Therefore, if the sum is 121, N² = 121, so N = 11.

I think that's solid. I can't see any mistakes in my reasoning. The key was remembering the formula for the sum of the first N odd numbers, which is N squared. Then solving for N gives the square root of 121, which is 11. Verified by adding the numbers step by step and also by using the arithmetic series formula. So I'm confident that N is 11.

**Final Answer**
The value of $ N $ is \boxed{11}.

To solve the problem, we are given that the sum of the first $ N $ positive odd numbers is 121, and we are to find the value of $ N $.

---

### Step 1: Understanding the Pattern

The sequence of the first few positive odd numbers is:

$$
1, 3, 5, 7, 9, \dots
$$

Let’s compute the sum of the first few terms:

- $ 1 = 1^2 $
- $ 1 + 3 = 4 = 2^2 $
- $ 1 + 3 + 5 = 9 = 3^2 $
- $ 1 + 3 + 5 + 7 = 16 = 4^2 $

This pattern suggests that the sum of the first $ N $ positive odd numbers is:

$$
\text{Sum} = N^2
$$

---

### Step 2: Applying the Formula

We are told that the sum is 121. Using the formula:

$$
N^2 = 121
$$

Taking the square root of both sides:

$$
N = \sqrt{121} = 11
$$

---

### Step 3: Verification

To ensure the result is correct, we can verify by summing the first 11 odd numbers:

$$
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121
$$

This confirms that the sum of the first 11 odd numbers is indeed 121.

---

### Final Answer

$$
\boxed{11}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of N where the sum of the first N positive odd numbers is 121. Hmm, let me think. I remember that there's a formula for the sum of the first N odd numbers. Wait, isn't it something like N squared? Let me check that.

Let me start by recalling some examples. The first odd number is 1. If N is 1, the sum is 1. According to the formula, 1 squared is 1. That works. The first two odd numbers are 1 and 3. Their sum is 4, which is 2 squared. The first three odd numbers: 1 + 3 + 5 = 9, which is 3 squared. Oh, right! So the pattern seems to be that the sum of the first N odd numbers is N². So if the sum is 121, then N² = 121. Therefore, N should be the square root of 121. Let me calculate that. The square root of 121 is 11, because 11 times 11 is 121. So N is 11? Let me verify this to make sure I didn't make a mistake.

Let me add up the first 11 odd numbers to check. The sequence of odd numbers is 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21. Let me add them step by step.

Starting with 1 + 3 = 4. Then 4 + 5 = 9. 9 + 7 = 16. 16 + 9 = 25. 25 + 11 = 36. 36 + 13 = 49. 49 + 15 = 64. 64 + 17 = 81. 81 + 19 = 100. 100 + 21 = 121. Wow, that adds up to 121. So N is indeed 11.

But just to make sure I didn't skip any steps or make an arithmetic error, let me try another way. Maybe using the formula for the sum of an arithmetic series. The sum of an arithmetic series is (number of terms)/2 times (first term + last term).

The first term is 1, and the Nth term is 2N - 1. So the sum would be N/2 * (1 + (2N - 1)) = N/2 * (2N) = N * N = N². Yep, that confirms the formula. So if the sum is 121, then N² = 121, so N = sqrt(121) = 11.

Alternatively, maybe I can think about the fact that each time you add an odd number, you're essentially building a square. Like, 1 is 1x1, then adding 3 gives 2x2, adding 5 gives 3x3, and so on. So the Nth odd number is the one that makes the square go from (N-1)² to N². Therefore, the sum of the first N odd numbers is N².

So, if the sum is 121, then N is 11.

Wait, but just to be thorough, let me check if there's any possibility that N could be negative? But since N is the number of terms, it has to be a positive integer. So N = 11 is the only solution.

Another way to think about it: if I didn't remember the formula, I could derive it. Let's say the sum S = 1 + 3 + 5 + ... + (2N - 1). Let me write this out:

S = 1 + 3 + 5 + ... + (2N - 1)

If I add the first term and the last term: 1 + (2N - 1) = 2N

Similarly, the second term and the second last term: 3 + (2N - 3) = 2N

And so on. Since there are N terms, there are N/2 such pairs if N is even, or (N-1)/2 pairs plus the middle term if N is odd. But in this case, since all the pairs add up to 2N, and there are N terms, which is either even or odd. Wait, if N is even, then there are N/2 pairs, each adding to 2N, so total sum is N/2 * 2N = N². If N is odd, then there are (N-1)/2 pairs adding to 2N, and the middle term is N. So total sum is (N-1)/2 * 2N + N = N(N - 1) + N = N² - N + N = N². So regardless of whether N is even or odd, the sum is N². Therefore, the formula holds.

Therefore, if the sum is 121, N² = 121, so N = 11.

I think that's solid. I can't see any mistakes in my reasoning. The key was remembering the formula for the sum of the first N odd numbers, which is N squared. Then solving for N gives the square root of 121, which is 11. Verified by adding the numbers step by step and also by using the arithmetic series formula. So I'm confident that N is 11.

**Final Answer**
The value of $ N $ is \boxed{11}.

To solve the problem, we are given that the sum of the first $ N $ positive odd numbers is 121, and we are to find the value of $ N $.

---

### Step 1: Understanding the Pattern

The sequence of the first few positive odd numbers is:

$$
1, 3, 5, 7, 9, \dots
$$

Let’s compute the sum of the first few terms:

- $ 1 = 1^2 $
- $ 1 + 3 = 4 = 2^2 $
- $ 1 + 3 + 5 = 9 = 3^2 $
- $ 1 + 3 + 5 + 7 = 16 = 4^2 $

This pattern suggests that the sum of the first $ N $ positive odd numbers is:

$$
\text{Sum} = N^2
$$

---

### Step 2: Applying the Formula

We are told that the sum is 121. Using the formula:

$$
N^2 = 121
$$

Taking the square root of both sides:

$$
N = \sqrt{121} = 11
$$

---

### Step 3: Verification

To ensure the result is correct, we can verify by summing the first 11 odd numbers:

$$
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 121
$$

This confirms that the sum of the first 11 odd numbers is indeed 121.

---

### Final Answer

$$
\boxed{11}
$$

Solution: The first $N$ positive odd integers are 1, 3, $\dots$, $2N - 1$.  The sum of an arithmetic series is equal to the average of the first and last term, multiplied by the number of terms, so the sum of the first $N$ positive odd integers is \[\frac{1 + (2N - 1)}{2} \cdot N = N^2.\]If $N^2 = 121$, then $N = \boxed{11}$.


Explanation:
Okay, so I need to find the value of $ a $ given that the cubic equation $ 8x^3 + 4ax^2 + 2bx + a = 0 $ has three distinct positive roots, and the sum of the base-2 logarithms of the roots is 5. Hmm, let me start by recalling some algebra concepts.

First, since the equation is a cubic with three distinct positive roots, I can use Vieta's formulas to relate the coefficients of the polynomial to the sums and products of its roots. Let me denote the roots as $ r $, $ s $, and $ t $. All of them are positive and distinct.

Vieta's formulas for a cubic equation $ px^3 + qx^2 + rx + s = 0 $ tell us that:

1. $ r + s + t = -q/p $
2. $ rs + rt + st = r/p $
3. $ rst = -s/p $

Wait, let me check that again. For a general cubic equation $ ax^3 + bx^2 + cx + d = 0 $, the sum of roots is $ -b/a $, the sum of products two at a time is $ c/a $, and the product of roots is $ -d/a $. So applying that to our equation:

Given the equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $, so here $ p = 8 $, $ q = 4a $, $ r = 2b $, and $ s = a $. Therefore:

1. Sum of roots: $ r + s + t = -q/p = -4a/8 = -a/2 $
2. Sum of products two at a time: $ rs + rt + st = r/p = 2b/8 = b/4 $
3. Product of roots: $ rst = -s/p = -a/8 $

But wait, the problem says the roots are positive. So the sum of the roots is $ -a/2 $. But since the roots are positive, their sum must be positive. Therefore, $ -a/2 > 0 $ implies that $ a < 0 $. So $ a $ is negative. Got that.

Also, the product of the roots is $ -a/8 $. Since the roots are positive, their product is positive, so $ -a/8 > 0 $, which again implies $ a < 0 $. So that's consistent.

Now, the problem says that the sum of the base-2 logarithms of the roots is 5. Let me write that down:

$ \log_2 r + \log_2 s + \log_2 t = 5 $

Using logarithm properties, this is equivalent to:

$ \log_2 (rst) = 5 $

Which means:

$ rst = 2^5 = 32 $

But from Vieta's formula, we have $ rst = -a/8 $. Therefore:

$ -a/8 = 32 $

Solving for $ a $:

Multiply both sides by 8:

$ -a = 256 $

Multiply both sides by -1:

$ a = -256 $

Wait, is that all? Let me check if there are any other conditions or if I missed something.

Wait, the problem mentions that the roots are distinct and positive. So, we need to make sure that with $ a = -256 $, the equation actually has three distinct positive roots. But maybe there's more to it? Let me think.

But the problem says "for some real numbers $ a $ and $ b $", so maybe there's a relation between $ a $ and $ b $ that we need to check? Wait, but the question only asks for $ a $, so maybe once we find $ a $, we can check if such $ b $ exists? But perhaps the given condition about the sum of logs already gives us $ a $, and the rest is just ensuring that the equation has three distinct positive roots. However, since the problem says "has three distinct positive roots", maybe there's more constraints?

Wait, but maybe I need to check if with $ a = -256 $, the equation can have three distinct positive roots. Let me think. Let me recall that for a cubic equation to have three distinct real roots, its discriminant must be positive. But calculating the discriminant of a cubic might be complicated. Alternatively, maybe we can use Vieta's formulas and other relations?

Alternatively, maybe there's another condition from the sum of the roots and the sum of products?

Wait, let me check again. The problem gives me that the sum of the logs is 5, which gives me $ rst = 32 $, and from Vieta's formula, $ rst = -a/8 $, so $ a = -256 $. But then, what about the other Vieta's formulas? For example, the sum of roots is $ -a/2 = -(-256)/2 = 128 $. So the sum of the roots is 128, and the product is 32. Also, the sum of products two at a time is $ b/4 $, so $ b = 4(rs + rt + st) $. But since the problem doesn't give us any information about $ b $, maybe we don't need to worry about it? The problem only asks for $ a $, so maybe even if there are multiple possible $ b $, as long as there exists some $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable?

But maybe there's something else? Let me think. For example, if the roots are positive, then the coefficients have to satisfy certain conditions. For example, the coefficients of the polynomial must be such that the polynomial can be factored with positive roots. But since we already used Vieta's formulas, and the sum and product, maybe that's sufficient?

Alternatively, maybe there's a mistake in my calculation? Let me check again:

Given $ \log_2 r + \log_2 s + \log_2 t = 5 $, so $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. Then Vieta's formula says $ rst = -a/8 $, so $ -a/8 = 32 $, so $ a = -256 $. That seems straightforward.

But let me check if there's another condition. For example, the sum of the roots is $ r + s + t = -a/2 = 128 $, and the product is 32. So, if we have three positive numbers with sum 128 and product 32, is that possible? Well, for example, if two roots are very small and one is almost 128, but then the product would be small. Wait, but 32 is quite small compared to 128. Wait, but maybe they are all fractions?

Wait, but if all roots are positive, then their product is 32, and their sum is 128. Let me think of an example. Suppose all three roots are equal? Then each would be 128/3 ≈ 42.666..., and the product would be (128/3)^3 ≈ 237, which is way larger than 32. So if they are equal, the product is too big. Therefore, to get a smaller product, the roots must be smaller. But how can they have a sum of 128 and product 32? Let me think.

Suppose two roots are very small, approaching zero, then the third root approaches 128, and the product approaches zero. So to get a product of 32, the roots can't be too small. Maybe there's a balance. Let me try to think of specific numbers. Let me suppose that two roots are equal? Wait, but the problem says three distinct roots. So they can't be equal. But even so, maybe there's a way to have three distinct positive numbers with sum 128 and product 32. For example, let me take one root as 1, another as 2, then the third root would be 128 - 1 - 2 = 125. Then the product is 1*2*125 = 250, which is still bigger than 32. If I take roots as 1, 1, 126, but they need to be distinct. Let me take 1, 2, 125 as before. If I take smaller numbers, say 0.5, 0.5, 127, but again, they need to be distinct. Let me take 0.5, 1, 126.5. The product is 0.5*1*126.5 = 63.25, still bigger than 32. If I take even smaller numbers, say 0.25, 0.25, 127.5, but again, distinctness. Let me take 0.25, 0.5, 127.25. Product is 0.25*0.5*127.25 ≈ 15.9, which is less than 32. So somewhere between 0.25 and 1, there must be some roots where the product is 32. So maybe there are such numbers. Therefore, it's possible. Therefore, even though the numbers seem counterintuitive, there could be three distinct positive numbers with sum 128 and product 32. Therefore, maybe $ a = -256 $ is acceptable.

But then why does the problem mention that there are three distinct positive roots? Maybe to ensure that the equation doesn't have multiple roots or negative roots? But since we already have the product positive and sum positive, and the coefficients are set accordingly, maybe that's sufficient?

Alternatively, maybe there's a mistake in assuming that Vieta's formula applies directly? Let me check again.

Original equation: $ 8x^3 + 4a x^2 + 2b x + a = 0 $. So if we write it as $ 8x^3 + 4a x^2 + 2b x + a = 0 $, then Vieta's formula for cubic equation $ px^3 + qx^2 + rx + s = 0 $:

Sum of roots: $ -q/p = -4a/8 = -a/2 $

Sum of products two at a time: $ r/p = 2b/8 = b/4 $

Product of roots: $ -s/p = -a/8 $

Yes, that's correct. So the product of roots is $ -a/8 $, which equals 32, so $ a = -256 $. Therefore, unless there's a miscalculation here, this should be correct.

But let me check if there's another condition. For example, maybe the equation must have three real roots? But the problem says three distinct positive roots, so they must be real. However, the problem didn't specify that the equation has three real roots, but since it says three distinct positive roots, they must be real. Therefore, maybe we need to ensure that with $ a = -256 $, the equation has three real roots. But how?

Alternatively, maybe the problem is designed so that once you satisfy the product and sum conditions, the other conditions are automatically satisfied? But I need to check if with $ a = -256 $, there exists a real number $ b $ such that the equation has three distinct positive roots.

Alternatively, maybe there's another relation between $ a $ and $ b $ that I haven't considered? For example, maybe the discriminant?

But calculating the discriminant of a cubic is complicated. Let me recall that for a cubic equation $ ax^3 + bx^2 + cx + d = 0 $, the discriminant $ D $ is given by:

$ D = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 $

But this seems complicated. However, since we know the roots, maybe we can use the discriminant in terms of roots. The discriminant is also equal to $ (r - s)^2 (s - t)^2 (t - r)^2 $. Since the roots are distinct, the discriminant must be positive. However, without knowing the specific values of the roots, it's hard to compute. But maybe with the given information, we can find $ b $?

Wait, we know that $ r + s + t = 128 $, $ rs + rt + st = b/4 $, and $ rst = 32 $. So if we can find $ b $, then we can check if the discriminant is positive. However, since the problem doesn't give us more information, maybe there's a way to find $ b $?

Alternatively, maybe there's another condition? Wait, the problem says "for some real numbers $ a $ and $ b $", so as long as there exists a real number $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable. Therefore, even if we don't know $ b $, as long as such $ b $ exists, then $ a = -256 $ is the answer.

But how can we be sure that such $ b $ exists? Well, given that we have three positive numbers $ r, s, t $ with sum 128 and product 32, then $ rs + rt + st $ is some positive number, so $ b = 4(rs + rt + st) $, which is real. Therefore, such $ b $ exists. Therefore, the answer is $ a = -256 $.

But let me check again if I made a mistake in the logarithm part. The problem says the sum of the base-2 logarithms of the roots is 5. So:

$ \log_2 r + \log_2 s + \log_2 t = 5 $

Which is $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. That's correct. Then Vieta's formula says $ rst = -a/8 $, so $ -a/8 = 32 $, so $ a = -256 $. That seems correct.

Wait, but let me check if the original equation with $ a = -256 $ would have three distinct positive roots. Let me try to think of an example. Suppose the roots are $ r, s, t $, with $ r + s + t = 128 $, $ rst = 32 $. Let me try to find such numbers. For example, suppose two roots are reciprocals? Not sure. Alternatively, maybe take one root as 32, and the other two roots as something?

Wait, if one root is 32, then the product of the other two roots is 32 / 32 = 1. Let me say the other two roots are $ x $ and $ 1/x $, so that their product is 1. Then the sum of all three roots is 32 + x + 1/x = 128. Therefore, x + 1/x = 128 - 32 = 96. Then x + 1/x = 96. Multiply both sides by x: x^2 - 96x + 1 = 0. The discriminant is 96^2 - 4*1*1 = 9216 - 4 = 9212, which is positive, so there are two real roots. Therefore, such numbers exist. Therefore, there are three distinct positive roots. Therefore, with $ a = -256 $, there exists such roots. Therefore, the answer is $ a = -256 $.

But let me check if there's a possible mistake in the sign. Let me check Vieta's formula again. The product of the roots is $ -a/8 $. Since the roots are positive, their product is positive, so $ -a/8 > 0 \implies a < 0 $. So with $ a = -256 $, that's correct. The sum of roots is $ -a/2 = 128 $, which is positive, as required. So that's all good.

Therefore, I think the answer is $ \boxed{-256} $.

Wait, but let me check once more. Maybe the problem says "three distinct positive roots", so maybe there's a condition on the coefficients? For example, the coefficients must satisfy certain inequalities? For example, if all roots are positive, then the coefficients must satisfy certain relations. For example, in a cubic equation with all positive roots, the coefficients must be positive or negative? Let me think.

Given the equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $. If all roots are positive, then by Vieta's formulas:

Sum of roots: $ -4a/8 = -a/2 > 0 \implies a < 0 $

Sum of products two at a time: $ 2b/8 = b/4 $. Since the roots are positive, the sum of products two at a time is positive, so $ b/4 > 0 \implies b > 0 $

Product of roots: $ -a/8 > 0 \implies a < 0 $, which is consistent.

Therefore, with $ a = -256 $, we have $ b > 0 $. Therefore, as long as $ b $ is positive, the equation can have three positive roots. Since the problem says "for some real numbers $ a $ and $ b $", so as long as there exists such $ b $, which there is, since $ b = 4(rs + rt + st) $, and $ rs + rt + st $ is positive, then $ b $ is positive. Therefore, all conditions are satisfied.

Therefore, I think my answer is correct. So the value of $ a $ is $ -256 $.

**Final Answer**
\boxed{-256}

To solve the given problem, we are told that the cubic equation

$$
8x^3 + 4ax^2 + 2bx + a = 0
$$

has **three distinct positive roots**, and that the **sum of the base-2 logarithms of the roots is 5**. We are to find the value of $ a $.

---

### Step 1: Use the Logarithmic Condition

Let the roots of the cubic be $ r, s, t $, all positive and distinct. The condition given is:

$$
\log_2 r + \log_2 s + \log_2 t = 5
$$

Using the logarithmic identity $ \log_b x + \log_b y + \log_b z = \log_b(xyz) $, this becomes:

$$
\log_2(rst) = 5 \Rightarrow rst = 2^5 = 32
$$

---

### Step 2: Apply Vieta's Formulas

From Vieta's formulas for a cubic equation of the form:

$$
px^3 + qx^2 + rx + s = 0
$$

we know:

- Sum of roots: $ r + s + t = -\frac{q}{p} $
- Sum of products of roots two at a time: $ rs + rt + st = \frac{r}{p} $
- Product of roots: $ rst = -\frac{s}{p} $

In our case, the cubic is:

$$
8x^3 + 4a x^2 + 2b x + a = 0
$$

So, $ p = 8 $, $ q = 4a $, $ r = 2b $, and $ s = a $. Applying Vieta's formulas:

- $ r + s + t = -\frac{4a}{8} = -\frac{a}{2} $
- $ rs + rt + st = \frac{2b}{8} = \frac{b}{4} $
- $ rst = -\frac{a}{8} $

We already found that $ rst = 32 $, so:

$$
-\frac{a}{8} = 32 \Rightarrow a = -256
$$

---

### Step 3: Verify Consistency with the Given Conditions

We now check whether this value of $ a $ is consistent with the rest of the problem.

- The product of the roots is $ rst = 32 $, which is positive, so the roots are all positive.
- The sum of the roots is $ r + s + t = -\frac{a}{2} = \frac{256}{2} = 128 $, which is also positive.
- The sum of the products of the roots two at a time is $ rs + rt + st = \frac{b}{4} $, which is positive as long as $ b > 0 $, and since the roots are positive, this is valid.

Thus, with $ a = -256 $, the equation has three distinct positive roots, and the given logarithmic condition is satisfied.

---

### Final Answer

$$
\boxed{-256}
$$

[Assistant]
Explanation:
Okay, so I need to find the value of $ a $ given that the cubic equation $ 8x^3 + 4ax^2 + 2bx + a = 0 $ has three distinct positive roots, and the sum of the base-2 logarithms of the roots is 5. Hmm, let me start by recalling some algebra concepts.

First, since the equation is a cubic with three distinct positive roots, I can use Vieta's formulas to relate the coefficients of the polynomial to the sums and products of its roots. Let me denote the roots as $ r $, $ s $, and $ t $. All of them are positive and distinct.

Vieta's formulas for a cubic equation $ px^3 + qx^2 + rx + s = 0 $ tell us that:

1. $ r + s + t = -q/p $
2. $ rs + rt + st = r/p $
3. $ rst = -s/p $

Wait, let me check that again. For a general cubic equation $ ax^3 + bx^2 + cx + d = 0 $, the sum of roots is $ -b/a $, the sum of products two at a time is $ c/a $, and the product of roots is $ -d/a $. So applying that to our equation:

Given the equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $, so here $ p = 8 $, $ q = 4a $, $ r = 2b $, and $ s = a $. Therefore:

1. Sum of roots: $ r + s + t = -q/p = -4a/8 = -a/2 $
2. Sum of products two at a time: $ rs + rt + st = r/p = 2b/8 = b/4 $
3. Product of roots: $ rst = -s/p = -a/8 $

But wait, the problem says the roots are positive. So the sum of the roots is $ -a/2 $. But since the roots are positive, their sum must be positive. Therefore, $ -a/2 > 0 $ implies that $ a < 0 $. So $ a $ is negative. Got that.

Also, the product of the roots is $ -a/8 $. Since the roots are positive, their product is positive, so $ -a/8 > 0 $, which again implies $ a < 0 $. So that's consistent.

Now, the problem says that the sum of the base-2 logarithms of the roots is 5. Let me write that down:

$ \log_2 r + \log_2 s + \log_2 t = 5 $

Using logarithm properties, this is equivalent to:

$ \log_2 (rst) = 5 $

Which means:

$ rst = 2^5 = 32 $

But from Vieta's formula, we have $ rst = -a/8 $. Therefore:

$ -a/8 = 32 $

Solving for $ a $:

Multiply both sides by 8:

$ -a = 256 $

Multiply both sides by -1:

$ a = -256 $

Wait, is that all? Let me check if there are any other conditions or if I missed something.

Wait, the problem mentions that the roots are distinct and positive. So, we need to make sure that with $ a = -256 $, the equation actually has three distinct positive roots. But maybe there's more to it? Let me think.

But the problem says "for some real numbers $ a $ and $ b $", so maybe there's a relation between $ a $ and $ b $ that we need to check? Wait, but the question only asks for $ a $, so maybe once we find $ a $, we can check if such $ b $ exists? But perhaps the given condition about the sum of logs already gives us $ a $, and the rest is just ensuring that the equation has three distinct positive roots. However, since the problem says "has three distinct positive roots", maybe there's more constraints?

Wait, but maybe I need to check if with $ a = -256 $, the equation can have three distinct positive roots. Let me think. Let me recall that for a cubic equation to have three distinct real roots, its discriminant must be positive. But calculating the discriminant of a cubic might be complicated. Alternatively, maybe we can use Vieta's formulas and other relations?

Alternatively, maybe there's another condition from the sum of the roots and the sum of products?

Wait, let me check again. The problem gives me that the sum of the logs is 5, which gives me $ rst = 32 $, and from Vieta's formula, $ rst = -a/8 $, so $ a = -256 $. But then, what about the other Vieta's formulas? For example, the sum of roots is $ -a/2 = -(-256)/2 = 128 $. So the sum of the roots is 128, and the product is 32. Also, the sum of products two at a time is $ b/4 $, so $ b = 4(rs + rt + st) $. But since the problem doesn't give us any information about $ b $, maybe we don't need to worry about it? The problem only asks for $ a $, so maybe even if there are multiple possible $ b $, as long as there exists some $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable?

But maybe there's something else? Let me think. For example, if the roots are positive, then the coefficients have to satisfy certain conditions. For example, the coefficients of the polynomial must be such that the polynomial can be factored with positive roots. But since we already used Vieta's formulas, and the sum and product, maybe that's sufficient?

Alternatively, maybe there's a mistake in my calculation? Let me check again:

Given $ \log_2 r + \log_2 s + \log_2 t = 5 $, so $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. Then Vieta's formula says $ rst = -a/8 $, so $ -a/8 = 32 $, so $ a = -256 $. That seems straightforward.

But let me check if there's another condition. For example, the sum of the roots is $ r + s + t = -a/2 = 128 $, and the product is 32. So, if we have three positive numbers with sum 128 and product 32, is that possible? Well, for example, if two roots are very small and one is almost 128, but then the product would be small. Wait, but 32 is quite small compared to 128. Wait, but maybe they are all fractions?

Wait, but if all roots are positive, then their product is 32, and their sum is 128. Let me think of an example. Suppose all three roots are equal? Then each would be 128/3 ≈ 42.666..., and the product would be (128/3)^3 ≈ 237, which is way larger than 32. So if they are equal, the product is too big. Therefore, to get a smaller product, the roots must be smaller. But how can they have a sum of 128 and product 32? Let me think.

Suppose two roots are very small, approaching zero, then the third root approaches 128, and the product approaches zero. So to get a product of 32, the roots can't be too small. Maybe there's a balance. Let me try to think of specific numbers. Let me suppose that two roots are equal? Wait, but the problem says three distinct roots. So they can't be equal. But even so, maybe there's a way to have three distinct positive numbers with sum 128 and product 32. For example, let me take one root as 1, another as 2, then the third root would be 128 - 1 - 2 = 125. Then the product is 1*2*125 = 250, which is still bigger than 32. If I take roots as 1, 1, 126, but they need to be distinct. Let me take 1, 2, 125 as before. If I take smaller numbers, say 0.5, 0.5, 127, but again, they need to be distinct. Let me take 0.5, 1, 126.5. The product is 0.5*1*126.5 = 63.25, still bigger than 32. If I take even smaller numbers, say 0.25, 0.25, 127.5, but again, distinctness. Let me take 0.25, 0.5, 127.25. Product is 0.25*0.5*127.25 ≈ 15.9, which is less than 32. So somewhere between 0.25 and 1, there must be some roots where the product is 32. So maybe there are such numbers. Therefore, it's possible. Therefore, even though the numbers seem counterintuitive, there could be three distinct positive numbers with sum 128 and product 32. Therefore, maybe $ a = -256 $ is acceptable.

But then why does the problem mention that there are three distinct positive roots? Maybe to ensure that the equation doesn't have multiple roots or negative roots? But since we already have the product positive and sum positive, and the coefficients are set accordingly, maybe that's sufficient?

Alternatively, maybe there's a mistake in assuming that Vieta's formula applies directly? Let me check again.

Original equation: $ 8x^3 + 4a x^2 + 2b x + a = 0 $. So if we write it as $ 8x^3 + 4a x^2 + 2b x + a = 0 $, then Vieta's formula for cubic equation $ px^3 + qx^2 + rx + s = 0 $:

Sum of roots: $ -q/p = -4a/8 = -a/2 $

Sum of products two at a time: $ r/p = 2b/8 = b/4 $

Product of roots: $ -s/p = -a/8 $

Yes, that's correct. So the product of roots is $ -a/8 $, which equals 32, so $ a = -256 $. Therefore, unless there's a miscalculation here, this should be correct.

But let me check if there's another condition. For example, maybe the equation must have three real roots? But the problem says three distinct positive roots, so they must be real. However, the problem didn't specify that the equation has three real roots, but since it says three distinct positive roots, they must be real. Therefore, maybe we need to ensure that with $ a = -256 $, the equation has three real roots. But how?

Alternatively, maybe the problem is designed so that once you satisfy the product and sum conditions, the other conditions are automatically satisfied? But I need to check if with $ a = -256 $, there exists a real number $ b $ such that the equation has three distinct positive roots.

Alternatively, maybe there's another relation between $ a $ and $ b $ that I haven't considered? For example, maybe the discriminant?

But calculating the discriminant of a cubic is complicated. Let me recall that for a cubic equation $ ax^3 + bx^2 + cx + d = 0 $, the discriminant $ D $ is given by:

$ D = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2 $

But this seems complicated. However, since we know the roots, maybe we can use the discriminant in terms of roots. The discriminant is also equal to $ (r - s)^2 (s - t)^2 (t - r)^2 $. Since the roots are distinct, the discriminant must be positive. However, without knowing the specific values of the roots, it's hard to compute. But maybe with the given information, we can find $ b $?

Wait, we know that $ r + s + t = 128 $, $ rs + rt + st = b/4 $, and $ rst = 32 $. So if we can find $ b $, then we can check if the discriminant is positive. However, since the problem doesn't give us more information, maybe there's a way to find $ b $?

Alternatively, maybe there's another condition? Wait, the problem says "for some real numbers $ a $ and $ b $", so as long as there exists a real number $ b $ such that the equation has three distinct positive roots, then $ a = -256 $ is acceptable. Therefore, even if we don't know $ b $, as long as such $ b $ exists, then $ a = -256 $ is the answer.

But how can we be sure that such $ b $ exists? Well, given that we have three positive numbers $ r, s, t $ with sum 128 and product 32, then $ rs + rt + st $ is some positive number, so $ b = 4(rs + rt + st) $, which is real. Therefore, such $ b $ exists. Therefore, the answer is $ a = -256 $.

But let me check again if I made a mistake in the logarithm part. The problem says the sum of the base-2 logarithms of the roots is 5. So:

$ \log_2 r + \log_2 s + \log_2 t = 5 $

Which is $ \log_2 (rst) = 5 $, so $ rst = 2^5 = 32 $. That's correct. Then Vieta's formula says $ rst = -a/8 $, so $ -a/8 = 32 $, so $ a = -256 $. That seems correct.

Wait, but let me check if the original equation with $ a = -256 $ would have three distinct positive roots. Let me try to think of an example. Suppose the roots are $ r, s, t $, with $ r + s + t = 128 $, $ rst = 32 $. Let me try to find such numbers. For example, suppose two roots are reciprocals? Not sure. Alternatively, maybe take one root as 32, and the other two roots as something?

Wait, if one root is 32, then the product of the other two roots is 32 / 32 = 1. Let me say the other two roots are $ x $ and $ 1/x $, so that their product is 1. Then the sum of all three roots is 32 + x + 1/x = 128. Therefore, x + 1/x = 128 - 32 = 96. Then x + 1/x = 96. Multiply both sides by x: x^2 - 96x + 1 = 0. The discriminant is 96^2 - 4*1*1 = 9216 - 4 = 9212, which is positive, so there are two real roots. Therefore, such numbers exist. Therefore, there are three distinct positive roots. Therefore, with $ a = -256 $, there exists such roots. Therefore, the answer is $ a = -256 $.

But let me check if there's a possible mistake in the sign. Let me check Vieta's formula again. The product of the roots is $ -a/8 $. Since the roots are positive, their product is positive, so $ -a/8 > 0 \implies a < 0 $. So with $ a = -256 $, that's correct. The sum of roots is $ -a/2 = 128 $, which is positive, as required. So that's all good.

Therefore, I think the answer is $ \boxed{-256} $.

Wait, but let me check once more. Maybe the problem says "three distinct positive roots", so maybe there's a condition on the coefficients? For example, the coefficients must satisfy certain inequalities? For example, if all roots are positive, then the coefficients must satisfy certain relations. For example, in a cubic equation with all positive roots, the coefficients must be positive or negative? Let me think.

Given the equation $ 8x^3 + 4a x^2 + 2b x + a = 0 $. If all roots are positive, then by Vieta's formulas:

Sum of roots: $ -4a/8 = -a/2 > 0 \implies a < 0 $

Sum of products two at a time: $ 2b/8 = b/4 $. Since the roots are positive, the sum of products two at a time is positive, so $ b/4 > 0 \implies b > 0 $

Product of roots: $ -a/8 > 0 \implies a < 0 $, which is consistent.

Therefore, with $ a = -256 $, we have $ b > 0 $. Therefore, as long as $ b $ is positive, the equation can have three positive roots. Since the problem says "for some real numbers $ a $ and $ b $", so as long as there exists such $ b $, which there is, since $ b = 4(rs + rt + st) $, and $ rs + rt + st $ is positive, then $ b $ is positive. Therefore, all conditions are satisfied.

Therefore, I think my answer is correct. So the value of $ a $ is $ -256 $.

**Final Answer**
\boxed{-256}

To solve the given problem, we are told that the cubic equation

$$
8x^3 + 4ax^2 + 2bx + a = 0
$$

has **three distinct positive roots**, and that the **sum of the base-2 logarithms of the roots is 5**. We are to find the value of $ a $.

---

### Step 1: Use the Logarithmic Condition

Let the roots of the cubic be $ r, s, t $, all positive and distinct. The condition given is:

$$
\log_2 r + \log_2 s + \log_2 t = 5
$$

Using the logarithmic identity $ \log_b x + \log_b y + \log_b z = \log_b(xyz) $, this becomes:

$$
\log_2(rst) = 5 \Rightarrow rst = 2^5 = 32
$$

---

### Step 2: Apply Vieta's Formulas

From Vieta's formulas for a cubic equation of the form:

$$
px^3 + qx^2 + rx + s = 0
$$

we know:

- Sum of roots: $ r + s + t = -\frac{q}{p} $
- Sum of products of roots two at a time: $ rs + rt + st = \frac{r}{p} $
- Product of roots: $ rst = -\frac{s}{p} $

In our case, the cubic is:

$$
8x^3 + 4a x^2 + 2b x + a = 0
$$

So, $ p = 8 $, $ q = 4a $, $ r = 2b $, and $ s = a $. Applying Vieta's formulas:

- $ r + s + t = -\frac{4a}{8} = -\frac{a}{2} $
- $ rs + rt + st = \frac{2b}{8} = \frac{b}{4} $
- $ rst = -\frac{a}{8} $

We already found that $ rst = 32 $, so:

$$
-\frac{a}{8} = 32 \Rightarrow a = -256
$$

---

### Step 3: Verify Consistency with the Given Conditions

We now check whether this value of $ a $ is consistent with the rest of the problem.

- The product of the roots is $ rst = 32 $, which is positive, so the roots are all positive.
- The sum of the roots is $ r + s + t = -\frac{a}{2} = \frac{256}{2} = 128 $, which is also positive.
- The sum of the products of the roots two at a time is $ rs + rt + st = \frac{b}{4} $, which is positive as long as $ b > 0 $, and since the roots are positive, this is valid.

Thus, with $ a = -256 $, the equation has three distinct positive roots, and the given logarithmic condition is satisfied.

---

### Final Answer

$$
\boxed{-256}
$$

Solution:
Let $r_1, r_2$, and $r_3$ be the roots. Then \[
5= \log_2r_1 + \log_2 r_2 + \log_2 r_3 = \log_2r_1r_2r_3,
\]so $r_1r_2r_3 = 2^5 = 32$. Since \[
8x^{3}+4ax^{2}+2bx+a=8(x-r_1)(x-r_2)(x-r_3),
\]it follows that $a = -8r_1r_2r_3= \boxed{-256}$.


Explanation:
Okay, so I need to expand and simplify this algebraic expression: x times (x times (1 + x) plus 2x) minus 3 times (x squared minus x plus 2). Let me take it step by step. First, I should probably start by looking at the innermost parentheses and work my way out.

Starting with the first part: x(1 + x). Wait, no, actually, the expression is x multiplied by [x(1 + x) + 2x]. So maybe I should first handle the part inside the brackets. Let me write that down.

So inside the brackets, there's x(1 + x) + 2x. Let me expand x(1 + x) first. Multiplying x by 1 gives x, and multiplying x by x gives x squared. So that part becomes x + x². Then we add 2x to that. So combining like terms, x + 2x is 3x. Therefore, the entire expression inside the brackets simplifies to x² + 3x.

Wait, let me check that again. Original inside the brackets: x(1 + x) + 2x. Expanding x(1 + x) is x*1 + x*x = x + x². Then adding 2x gives x + x² + 2x. Combining x and 2x gives 3x, so yes, it's x² + 3x. Okay, that seems right.

So now the original expression becomes x multiplied by (x² + 3x) minus 3 times (x² - x + 2). Let me write that down:

x(x² + 3x) - 3(x² - x + 2)

Now I need to expand both of these terms. Let me start with the first term: x multiplied by (x² + 3x). Distribute the x to each term inside the parentheses.

Multiplying x by x² gives x³, and multiplying x by 3x gives 3x². So the first part becomes x³ + 3x².

Now the second part is -3 multiplied by (x² - x + 2). Let me distribute the -3 to each term inside the parentheses.

First term: -3 * x² = -3x²

Second term: -3 * (-x) = +3x

Third term: -3 * 2 = -6

So putting that together, the second part is -3x² + 3x - 6.

Now, combining both parts: the first part was x³ + 3x² and the second part is -3x² + 3x - 6. So adding them together:

x³ + 3x² - 3x² + 3x - 6

Now let me combine like terms. The 3x² and -3x² cancel each other out because 3x² - 3x² is 0. So that leaves us with x³ + 3x - 6.

Wait, is that all? Let me check again.

Original expression after expansion:

First part: x³ + 3x²

Second part: -3x² + 3x - 6

Adding them together:

x³ + 3x² - 3x² + 3x - 6

Yes, the 3x² and -3x² cancel. So we have x³ + 3x - 6.

Is there anything else to simplify? Let me check if there are any more like terms. The terms are x³, 3x, and -6. There are no other like terms. So that should be the simplified form.

Wait, but let me go back through each step again to make sure I didn't make a mistake.

Starting with the original expression:

x[ x(1 + x) + 2x ] - 3(x² - x + 2)

First, inside the brackets: x(1 + x) + 2x.

x(1 + x) is x + x². Then adding 2x gives x + x² + 2x = x² + 3x. Correct.

Then multiplying by x: x*(x² + 3x) = x³ + 3x². Correct.

Then subtracting 3*(x² - x + 2). Distribute the -3: -3x² + 3x -6. Correct.

Combine with the previous terms: x³ + 3x² -3x² + 3x -6. The 3x² -3x² cancels. So x³ + 3x -6.

Yes, that seems right. Let me check if I can factor anything or if there's a different way to write it, but the question says to expand and simplify completely. Since there are no like terms left, this should be the final answer.

Alternatively, maybe I can check by plugging in a value for x. Let me test with x = 1.

Original expression:

First compute the original expression at x = 1:

1*(1*(1 + 1) + 2*1) - 3*(1² - 1 + 2)

Compute step by step:

Inside the first bracket: 1*(1*(2) + 2) = 1*(2 + 2) = 1*4 = 4

Then subtract 3*(1 -1 + 2) = 3*(0 + 2) = 3*2 = 6

So total expression is 4 - 6 = -2

Now compute my simplified expression x³ + 3x -6 at x = 1:

1³ + 3*1 -6 = 1 + 3 -6 = -2. Same result. Good.

Let me try another value, say x = 2.

Original expression:

2*(2*(1 + 2) + 2*2) - 3*(2² - 2 + 2)

Compute step by step:

Inside first bracket: 2*(2*3 + 4) = 2*(6 + 4) = 2*10 = 20

Subtract 3*(4 - 2 + 2) = 3*(4) = 12

Total expression: 20 - 12 = 8

Simplified expression: 2³ + 3*2 -6 = 8 + 6 -6 = 8. Correct again.

Another test, x = 0.

Original expression:

0*(0*(1 + 0) + 2*0) - 3*(0 - 0 + 2) = 0*(0 + 0) - 3*(2) = 0 - 6 = -6

Simplified expression: 0³ + 3*0 -6 = 0 + 0 -6 = -6. Correct.

Okay, so seems like my simplified expression is correct. Therefore, the answer is x³ + 3x -6.

**Final Answer**
The simplified expression is \boxed{x^3 + 3x - 6}.

To simplify the expression:

$$
x\left(x(1+x)+2x\right)-3(x^2-x+2)
$$

we proceed step by step, focusing on expanding and combining like terms.

---

### **Step 1: Simplify the inner expression**

Start with the innermost parentheses:

$$
x(1 + x) = x \cdot 1 + x \cdot x = x + x^2
$$

Now add the remaining term inside the brackets:

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

So the expression becomes:

$$
x(x^2 + 3x) - 3(x^2 - x + 2)
$$

---

### **Step 2: Expand each term**

**First term:**

$$
x(x^2 + 3x) = x \cdot x^2 + x \cdot 3x = x^3 + 3x^2
$$

**Second term:**

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

---

### **Step 3: Combine all terms**

Now combine the expanded parts:

$$
x^3 + 3x^2 - 3x^2 + 3x - 6
$$

Group and simplify like terms:

- $3x^2 - 3x^2 = 0$
- Remaining terms: $x^3 + 3x - 6$

---

### **Final Simplified Expression**

$$
\boxed{x^3 + 3x - 6}
$$

[Assistant]
Explanation:
Okay, so I need to expand and simplify this algebraic expression: x times (x times (1 + x) plus 2x) minus 3 times (x squared minus x plus 2). Let me take it step by step. First, I should probably start by looking at the innermost parentheses and work my way out.

Starting with the first part: x(1 + x). Wait, no, actually, the expression is x multiplied by [x(1 + x) + 2x]. So maybe I should first handle the part inside the brackets. Let me write that down.

So inside the brackets, there's x(1 + x) + 2x. Let me expand x(1 + x) first. Multiplying x by 1 gives x, and multiplying x by x gives x squared. So that part becomes x + x². Then we add 2x to that. So combining like terms, x + 2x is 3x. Therefore, the entire expression inside the brackets simplifies to x² + 3x.

Wait, let me check that again. Original inside the brackets: x(1 + x) + 2x. Expanding x(1 + x) is x*1 + x*x = x + x². Then adding 2x gives x + x² + 2x. Combining x and 2x gives 3x, so yes, it's x² + 3x. Okay, that seems right.

So now the original expression becomes x multiplied by (x² + 3x) minus 3 times (x² - x + 2). Let me write that down:

x(x² + 3x) - 3(x² - x + 2)

Now I need to expand both of these terms. Let me start with the first term: x multiplied by (x² + 3x). Distribute the x to each term inside the parentheses.

Multiplying x by x² gives x³, and multiplying x by 3x gives 3x². So the first part becomes x³ + 3x².

Now the second part is -3 multiplied by (x² - x + 2). Let me distribute the -3 to each term inside the parentheses.

First term: -3 * x² = -3x²

Second term: -3 * (-x) = +3x

Third term: -3 * 2 = -6

So putting that together, the second part is -3x² + 3x - 6.

Now, combining both parts: the first part was x³ + 3x² and the second part is -3x² + 3x - 6. So adding them together:

x³ + 3x² - 3x² + 3x - 6

Now let me combine like terms. The 3x² and -3x² cancel each other out because 3x² - 3x² is 0. So that leaves us with x³ + 3x - 6.

Wait, is that all? Let me check again.

Original expression after expansion:

First part: x³ + 3x²

Second part: -3x² + 3x - 6

Adding them together:

x³ + 3x² - 3x² + 3x - 6

Yes, the 3x² and -3x² cancel. So we have x³ + 3x - 6.

Is there anything else to simplify? Let me check if there are any more like terms. The terms are x³, 3x, and -6. There are no other like terms. So that should be the simplified form.

Wait, but let me go back through each step again to make sure I didn't make a mistake.

Starting with the original expression:

x[ x(1 + x) + 2x ] - 3(x² - x + 2)

First, inside the brackets: x(1 + x) + 2x.

x(1 + x) is x + x². Then adding 2x gives x + x² + 2x = x² + 3x. Correct.

Then multiplying by x: x*(x² + 3x) = x³ + 3x². Correct.

Then subtracting 3*(x² - x + 2). Distribute the -3: -3x² + 3x -6. Correct.

Combine with the previous terms: x³ + 3x² -3x² + 3x -6. The 3x² -3x² cancels. So x³ + 3x -6.

Yes, that seems right. Let me check if I can factor anything or if there's a different way to write it, but the question says to expand and simplify completely. Since there are no like terms left, this should be the final answer.

Alternatively, maybe I can check by plugging in a value for x. Let me test with x = 1.

Original expression:

First compute the original expression at x = 1:

1*(1*(1 + 1) + 2*1) - 3*(1² - 1 + 2)

Compute step by step:

Inside the first bracket: 1*(1*(2) + 2) = 1*(2 + 2) = 1*4 = 4

Then subtract 3*(1 -1 + 2) = 3*(0 + 2) = 3*2 = 6

So total expression is 4 - 6 = -2

Now compute my simplified expression x³ + 3x -6 at x = 1:

1³ + 3*1 -6 = 1 + 3 -6 = -2. Same result. Good.

Let me try another value, say x = 2.

Original expression:

2*(2*(1 + 2) + 2*2) - 3*(2² - 2 + 2)

Compute step by step:

Inside first bracket: 2*(2*3 + 4) = 2*(6 + 4) = 2*10 = 20

Subtract 3*(4 - 2 + 2) = 3*(4) = 12

Total expression: 20 - 12 = 8

Simplified expression: 2³ + 3*2 -6 = 8 + 6 -6 = 8. Correct again.

Another test, x = 0.

Original expression:

0*(0*(1 + 0) + 2*0) - 3*(0 - 0 + 2) = 0*(0 + 0) - 3*(2) = 0 - 6 = -6

Simplified expression: 0³ + 3*0 -6 = 0 + 0 -6 = -6. Correct.

Okay, so seems like my simplified expression is correct. Therefore, the answer is x³ + 3x -6.

**Final Answer**
The simplified expression is \boxed{x^3 + 3x - 6}.

To simplify the expression:

$$
x\left(x(1+x)+2x\right)-3(x^2-x+2)
$$

we proceed step by step, focusing on expanding and combining like terms.

---

### **Step 1: Simplify the inner expression**

Start with the innermost parentheses:

$$
x(1 + x) = x \cdot 1 + x \cdot x = x + x^2
$$

Now add the remaining term inside the brackets:

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

So the expression becomes:

$$
x(x^2 + 3x) - 3(x^2 - x + 2)
$$

---

### **Step 2: Expand each term**

**First term:**

$$
x(x^2 + 3x) = x \cdot x^2 + x \cdot 3x = x^3 + 3x^2
$$

**Second term:**

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

---

### **Step 3: Combine all terms**

Now combine the expanded parts:

$$
x^3 + 3x^2 - 3x^2 + 3x - 6
$$

Group and simplify like terms:

- $3x^2 - 3x^2 = 0$
- Remaining terms: $x^3 + 3x - 6$

---

### **Final Simplified Expression**

$$
\boxed{x^3 + 3x - 6}
$$

Solution:
Begin by distributing in the innermost parentheses: \begin{align*}
&\ \ \ \ x\left(x(1+x)+2x\right)-3(x^2-x+2) \\&= x(x+x^2+2x) - 3(x^2-x+2)
\end{align*} Now, distribute again: \begin{align*}
x^2+x^3+2x^2-3x^2+3x-6
\end{align*} Finally, combine like terms to get \begin{align*}
\boxed{x^3+3x-6}
\end{align*}


Explanation:
Okay, let's see. I need to figure out how many treeks weigh the same as one squig. The problem gives me two equations, so I should probably set up some variables and solve them. Let me start by assigning variables to each of the creatures. Let me call the weight of a treek "T", a squig "S", and a goolee "G".

The first statement says: Ten treeks weigh as much as three squigs and one goolee. So in equation form, that would be 10T = 3S + G. Got that.

The second statement says: Two treeks and one goolee are equal in weight to one squig. So that would be 2T + G = S.

So now I have two equations:

1) 10T = 3S + G

2) 2T + G = S

And the question is asking: The combined weight of how many treeks equals the weight of one squig? So they want to know what number multiplied by T equals S. In other words, find x such that xT = S. So x is the number of treeks equivalent to one squig.

Alright, so I need to solve these equations to find S in terms of T, or maybe find the ratio between S and T. Let me see. Since equation 2 gives me S in terms of T and G, maybe I can substitute that into equation 1.

From equation 2: S = 2T + G. So I can replace S in equation 1 with that expression. Let me do that.

Substituting into equation 1: 10T = 3*(2T + G) + G. Let me compute the right side.

First, 3*(2T + G) is 6T + 3G. Then adding the G from the original equation gives 6T + 3G + G = 6T + 4G.

So equation 1 becomes: 10T = 6T + 4G.

Now, subtract 6T from both sides: 10T - 6T = 4G => 4T = 4G.

Divide both sides by 4: T = G.

Oh! So a treek weighs the same as a goolee? Interesting. So T = G.

Now, going back to equation 2: S = 2T + G. But since G = T, substitute that in: S = 2T + T = 3T.

So S = 3T. Therefore, one squig weighs the same as three treeks.

Wait, so the combined weight of three treeks equals the weight of one squig. Therefore, the answer is 3?

Let me check my steps again to make sure I didn't make a mistake.

Starting with equations:

1) 10T = 3S + G

2) 2T + G = S

Substituted equation 2 into equation 1:

10T = 3*(2T + G) + G => 10T = 6T + 3G + G => 10T = 6T + 4G => 4T = 4G => T = G.

Then equation 2: S = 2T + G = 2T + T = 3T.

Yes, that seems correct. So S = 3T. Therefore, one squig is equal to three treeks.

But let me verify with the original equations. Let's check if these values satisfy both equations.

First equation: 10T = 3S + G. If S = 3T and G = T, then the right side is 3*(3T) + T = 9T + T = 10T. Which equals the left side. Good.

Second equation: 2T + G = S. Substituting, 2T + T = 3T = S. Which is correct.

So both equations are satisfied. Therefore, the answer is 3.

But wait, the problem says "the combined weight of how many treeks equals the weight of one squig?" So that would be 3 treeks.

But let me think again. Maybe I misread the problem? Let me check again.

Original problem:

Ten treeks weigh as much as three squigs and one goolee.

So 10T = 3S + G.

Two treeks and one goolee are equal in weight to one squig.

So 2T + G = S.

Yes, that's what I used.

So solving gives S = 3T. Therefore, 3 treeks equal one squig.

Therefore, the answer is 3.

But just to make sure, let me try plugging in numbers. Let me assign a value to T. Let's say T = 1 (treek weighs 1 unit). Then G = T = 1. Then S = 3T = 3.

Check first equation: 10T = 10*1 = 10.

3S + G = 3*3 + 1 = 9 + 1 = 10. Correct.

Second equation: 2T + G = 2*1 + 1 = 3, which is equal to S. Correct.

So yes, that works. Therefore, the answer is 3.

I think that's solid. So the combined weight of three treeks equals one squig.

**Final Answer**
The combined weight of \boxed{3} treeks equals the weight of one squig.

To determine how many treeks weigh the same as one squig, we begin by translating the given information into algebraic equations.

---

### Step 1: Define Variables

Let:
- $ T $ = weight of one treek
- $ S $ = weight of one squig
- $ G $ = weight of one goolee

---

### Step 2: Translate the Problem into Equations

From the problem, we are given two key relationships:

1. **Ten treeks weigh as much as three squigs and one goolee:**
   $$
   10T = 3S + G \tag{1}
   $$

2. **Two treeks and one goolee are equal in weight to one squig:**
   $$
   2T + G = S \tag{2}
   $$

---

### Step 3: Solve the System of Equations

From equation (2), solve for $ S $ in terms of $ T $ and $ G $:
$$
S = 2T + G \tag{3}
$$

Now substitute equation (3) into equation (1):
$$
10T = 3(2T + G) + G
$$
$$
10T = 6T + 3G + G
$$
$$
10T = 6T + 4G
$$
$$
4T = 4G \Rightarrow T = G \tag{4}
$$

Now substitute $ T = G $ into equation (3):
$$
S = 2T + T = 3T
$$

---

### Step 4: Interpret the Result

We now know:
$$
S = 3T
$$

This means **one squig weighs the same as three treeks**.

---

### Final Answer

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, let's see. I need to figure out how many treeks weigh the same as one squig. The problem gives me two equations, so I should probably set up some variables and solve them. Let me start by assigning variables to each of the creatures. Let me call the weight of a treek "T", a squig "S", and a goolee "G".

The first statement says: Ten treeks weigh as much as three squigs and one goolee. So in equation form, that would be 10T = 3S + G. Got that.

The second statement says: Two treeks and one goolee are equal in weight to one squig. So that would be 2T + G = S.

So now I have two equations:

1) 10T = 3S + G

2) 2T + G = S

And the question is asking: The combined weight of how many treeks equals the weight of one squig? So they want to know what number multiplied by T equals S. In other words, find x such that xT = S. So x is the number of treeks equivalent to one squig.

Alright, so I need to solve these equations to find S in terms of T, or maybe find the ratio between S and T. Let me see. Since equation 2 gives me S in terms of T and G, maybe I can substitute that into equation 1.

From equation 2: S = 2T + G. So I can replace S in equation 1 with that expression. Let me do that.

Substituting into equation 1: 10T = 3*(2T + G) + G. Let me compute the right side.

First, 3*(2T + G) is 6T + 3G. Then adding the G from the original equation gives 6T + 3G + G = 6T + 4G.

So equation 1 becomes: 10T = 6T + 4G.

Now, subtract 6T from both sides: 10T - 6T = 4G => 4T = 4G.

Divide both sides by 4: T = G.

Oh! So a treek weighs the same as a goolee? Interesting. So T = G.

Now, going back to equation 2: S = 2T + G. But since G = T, substitute that in: S = 2T + T = 3T.

So S = 3T. Therefore, one squig weighs the same as three treeks.

Wait, so the combined weight of three treeks equals the weight of one squig. Therefore, the answer is 3?

Let me check my steps again to make sure I didn't make a mistake.

Starting with equations:

1) 10T = 3S + G

2) 2T + G = S

Substituted equation 2 into equation 1:

10T = 3*(2T + G) + G => 10T = 6T + 3G + G => 10T = 6T + 4G => 4T = 4G => T = G.

Then equation 2: S = 2T + G = 2T + T = 3T.

Yes, that seems correct. So S = 3T. Therefore, one squig is equal to three treeks.

But let me verify with the original equations. Let's check if these values satisfy both equations.

First equation: 10T = 3S + G. If S = 3T and G = T, then the right side is 3*(3T) + T = 9T + T = 10T. Which equals the left side. Good.

Second equation: 2T + G = S. Substituting, 2T + T = 3T = S. Which is correct.

So both equations are satisfied. Therefore, the answer is 3.

But wait, the problem says "the combined weight of how many treeks equals the weight of one squig?" So that would be 3 treeks.

But let me think again. Maybe I misread the problem? Let me check again.

Original problem:

Ten treeks weigh as much as three squigs and one goolee.

So 10T = 3S + G.

Two treeks and one goolee are equal in weight to one squig.

So 2T + G = S.

Yes, that's what I used.

So solving gives S = 3T. Therefore, 3 treeks equal one squig.

Therefore, the answer is 3.

But just to make sure, let me try plugging in numbers. Let me assign a value to T. Let's say T = 1 (treek weighs 1 unit). Then G = T = 1. Then S = 3T = 3.

Check first equation: 10T = 10*1 = 10.

3S + G = 3*3 + 1 = 9 + 1 = 10. Correct.

Second equation: 2T + G = 2*1 + 1 = 3, which is equal to S. Correct.

So yes, that works. Therefore, the answer is 3.

I think that's solid. So the combined weight of three treeks equals one squig.

**Final Answer**
The combined weight of \boxed{3} treeks equals the weight of one squig.

To determine how many treeks weigh the same as one squig, we begin by translating the given information into algebraic equations.

---

### Step 1: Define Variables

Let:
- $ T $ = weight of one treek
- $ S $ = weight of one squig
- $ G $ = weight of one goolee

---

### Step 2: Translate the Problem into Equations

From the problem, we are given two key relationships:

1. **Ten treeks weigh as much as three squigs and one goolee:**
   $$
   10T = 3S + G \tag{1}
   $$

2. **Two treeks and one goolee are equal in weight to one squig:**
   $$
   2T + G = S \tag{2}
   $$

---

### Step 3: Solve the System of Equations

From equation (2), solve for $ S $ in terms of $ T $ and $ G $:
$$
S = 2T + G \tag{3}
$$

Now substitute equation (3) into equation (1):
$$
10T = 3(2T + G) + G
$$
$$
10T = 6T + 3G + G
$$
$$
10T = 6T + 4G
$$
$$
4T = 4G \Rightarrow T = G \tag{4}
$$

Now substitute $ T = G $ into equation (3):
$$
S = 2T + T = 3T
$$

---

### Step 4: Interpret the Result

We now know:
$$
S = 3T
$$

This means **one squig weighs the same as three treeks**.

---

### Final Answer

$$
\boxed{3}
$$

Solution:
Let $t,s,g$ be the weight of one treek, the weight of one squig, and the weight of one goolee, respectively. Then the given information tells us \begin{align*}
10t &=3s+g\\
2t +g &= s.
\end{align*} Since we would like to solve for $s$ in terms of $t$, we want to eliminate $g$. Add the two equations to obtain \begin{align*}
10t+2t+g &= 3s+g+s\\
\Rightarrow 10t+2t &= 3s+s\\
\Rightarrow 4s &= 12t\\
\Rightarrow s &=3t.
\end{align*} So one squig weighs $\boxed{3}$ treeks.


Explanation:
Okay, so Remmy wants to divide 10 by 2/3, but he's confused about how to do that. Let me think about how to explain this step by step.

First, I remember that dividing by a fraction is the same as multiplying by its reciprocal. Wait, is that right? Let me make sure. Yeah, I think that's the rule. So if you have a division problem like a divided by b/c, you can flip the fraction b/c to c/b and then multiply. So instead of dividing by 2/3, you multiply by 3/2.

But let me verify this with an example. Let's say I have 2 divided by 1/2. If I follow the rule, that should be 2 multiplied by 2/1, which is 4. And indeed, 2 divided by 1/2 is 4 because there are four halves in 2. So that works. So maybe the same applies here.

So applying that to Remmy's problem: dividing 10 by 2/3. So instead of dividing by 2/3, he should multiply by 3/2. Therefore, the number he should multiply 10 by is 3/2.

But let me think again. Maybe I should write it out as a mathematical expression.

Dividing by a fraction is equivalent to multiplying by its reciprocal. So:

10 ÷ (2/3) = 10 × (3/2)

Yes, that seems correct. Let me check with another example. If I divide 6 by 2/3, that should be 6 × 3/2 = 9. Let me check that. 2/3 goes into 6 how many times? Well, 2/3 times 9 is 6, so yes, 9. So that works.

Alternatively, maybe I can think of division as asking "how many times does 2/3 fit into 10?" So if I have 10 and I want to know how many 2/3s are in there, multiplying by the reciprocal gives me that count.

Alternatively, maybe I can think of fractions. Dividing by a fraction is like multiplying by its reciprocal because when you divide by a fraction, you're essentially scaling the number by the inverse.

But let me also think about the mathematical reasoning behind it. Let's say we have a ÷ (b/c). Let me write that as a fraction: a divided by (b/c) is equal to a multiplied by c/b.

So, in this case, a is 10, b is 2, and c is 3. Therefore, 10 multiplied by 3/2.

Alternatively, if I want to do this without the reciprocal rule, maybe I can convert the division into a multiplication by using the concept of multiplying numerator and denominator.

So, dividing by 2/3 is the same as multiplying by 1/(2/3). And 1 divided by (2/3) is 3/2. Therefore, 10 multiplied by 3/2.

So, regardless of the method, it seems like the answer is multiplying by 3/2.

But let me also think about the actual calculation. If I do 10 divided by 2/3, that's the same as 10 multiplied by 3 divided by 2. So 10 times 3 is 30, divided by 2 is 15. So the result is 15.

Alternatively, if I multiply 10 by 3/2, that's (10 × 3)/2 = 30/2 = 15. So that's the same answer.

Therefore, the number he should multiply 10 by is 3/2.

But let me check if there's another way someone might approach this problem. Maybe by converting the division into a fraction.

So, 10 divided by 2/3 can be written as 10/(2/3). To simplify this complex fraction, you can multiply the numerator and denominator by 3 to eliminate the fraction in the denominator.

So, multiplying numerator and denominator by 3: (10 × 3)/( (2/3) × 3 ) = 30 / 2 = 15.

So again, we end up with 15, which is the same as multiplying 10 by 3/2.

Therefore, regardless of the method, the key step is multiplying by the reciprocal of the divisor.

Another way to think about it is using common denominators. If I want to divide 10 by 2/3, I can convert 10 into thirds. Since 10 is equal to 30/3, then dividing 30/3 by 2/3 is the same as (30/3) ÷ (2/3).

When you divide fractions with the same denominator, you can divide the numerators. So 30 divided by 2 is 15, and the denominators are both 3, so 3 divided by 3 is 1. Therefore, 15/1 = 15.

So again, this shows that dividing by 2/3 is equivalent to multiplying by 3/2.

Therefore, all these different methods lead to the conclusion that multiplying by 3/2 is the correct operation.

I think that's thorough enough. I can't think of any other way to approach this problem. All the different methods confirm that the reciprocal is the key here. So Remmy should multiply 10 by 3/2 to get the answer.

**Final Answer**
Remmy should multiply $10$ by \boxed{\dfrac{3}{2}} to get the answer.

To solve the problem of dividing $10$ by $\frac{2}{3}$, we need to understand the mathematical rule for dividing by a fraction.

---

### Step 1: Understanding the Rule

When you divide a number by a fraction, you are essentially asking: *How many times does the fraction fit into the number?* This is equivalent to **multiplying the number by the reciprocal of the fraction**.

The **reciprocal** of a fraction is obtained by **flipping the numerator and the denominator**. So, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

---

### Step 2: Applying the Rule

Given the expression:

$$
10 \div \frac{2}{3}
$$

We apply the rule:

$$
10 \div \frac{2}{3} = 10 \times \frac{3}{2}
$$

---

### Step 3: Performing the Multiplication

Now, compute the multiplication:

$$
10 \times \frac{3}{2} = \frac{10 \times 3}{2} = \frac{30}{2} = 15
$$

So, the result of dividing $10$ by $\frac{2}{3}$ is $15$.

---

### Final Answer

To get the answer, Remmy should multiply $10$ by the reciprocal of $\frac{2}{3}$, which is:

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

[Assistant]
Explanation:
Okay, so Remmy wants to divide 10 by 2/3, but he's confused about how to do that. Let me think about how to explain this step by step.

First, I remember that dividing by a fraction is the same as multiplying by its reciprocal. Wait, is that right? Let me make sure. Yeah, I think that's the rule. So if you have a division problem like a divided by b/c, you can flip the fraction b/c to c/b and then multiply. So instead of dividing by 2/3, you multiply by 3/2.

But let me verify this with an example. Let's say I have 2 divided by 1/2. If I follow the rule, that should be 2 multiplied by 2/1, which is 4. And indeed, 2 divided by 1/2 is 4 because there are four halves in 2. So that works. So maybe the same applies here.

So applying that to Remmy's problem: dividing 10 by 2/3. So instead of dividing by 2/3, he should multiply by 3/2. Therefore, the number he should multiply 10 by is 3/2.

But let me think again. Maybe I should write it out as a mathematical expression.

Dividing by a fraction is equivalent to multiplying by its reciprocal. So:

10 ÷ (2/3) = 10 × (3/2)

Yes, that seems correct. Let me check with another example. If I divide 6 by 2/3, that should be 6 × 3/2 = 9. Let me check that. 2/3 goes into 6 how many times? Well, 2/3 times 9 is 6, so yes, 9. So that works.

Alternatively, maybe I can think of division as asking "how many times does 2/3 fit into 10?" So if I have 10 and I want to know how many 2/3s are in there, multiplying by the reciprocal gives me that count.

Alternatively, maybe I can think of fractions. Dividing by a fraction is like multiplying by its reciprocal because when you divide by a fraction, you're essentially scaling the number by the inverse.

But let me also think about the mathematical reasoning behind it. Let's say we have a ÷ (b/c). Let me write that as a fraction: a divided by (b/c) is equal to a multiplied by c/b.

So, in this case, a is 10, b is 2, and c is 3. Therefore, 10 multiplied by 3/2.

Alternatively, if I want to do this without the reciprocal rule, maybe I can convert the division into a multiplication by using the concept of multiplying numerator and denominator.

So, dividing by 2/3 is the same as multiplying by 1/(2/3). And 1 divided by (2/3) is 3/2. Therefore, 10 multiplied by 3/2.

So, regardless of the method, it seems like the answer is multiplying by 3/2.

But let me also think about the actual calculation. If I do 10 divided by 2/3, that's the same as 10 multiplied by 3 divided by 2. So 10 times 3 is 30, divided by 2 is 15. So the result is 15.

Alternatively, if I multiply 10 by 3/2, that's (10 × 3)/2 = 30/2 = 15. So that's the same answer.

Therefore, the number he should multiply 10 by is 3/2.

But let me check if there's another way someone might approach this problem. Maybe by converting the division into a fraction.

So, 10 divided by 2/3 can be written as 10/(2/3). To simplify this complex fraction, you can multiply the numerator and denominator by 3 to eliminate the fraction in the denominator.

So, multiplying numerator and denominator by 3: (10 × 3)/( (2/3) × 3 ) = 30 / 2 = 15.

So again, we end up with 15, which is the same as multiplying 10 by 3/2.

Therefore, regardless of the method, the key step is multiplying by the reciprocal of the divisor.

Another way to think about it is using common denominators. If I want to divide 10 by 2/3, I can convert 10 into thirds. Since 10 is equal to 30/3, then dividing 30/3 by 2/3 is the same as (30/3) ÷ (2/3).

When you divide fractions with the same denominator, you can divide the numerators. So 30 divided by 2 is 15, and the denominators are both 3, so 3 divided by 3 is 1. Therefore, 15/1 = 15.

So again, this shows that dividing by 2/3 is equivalent to multiplying by 3/2.

Therefore, all these different methods lead to the conclusion that multiplying by 3/2 is the correct operation.

I think that's thorough enough. I can't think of any other way to approach this problem. All the different methods confirm that the reciprocal is the key here. So Remmy should multiply 10 by 3/2 to get the answer.

**Final Answer**
Remmy should multiply $10$ by \boxed{\dfrac{3}{2}} to get the answer.

To solve the problem of dividing $10$ by $\frac{2}{3}$, we need to understand the mathematical rule for dividing by a fraction.

---

### Step 1: Understanding the Rule

When you divide a number by a fraction, you are essentially asking: *How many times does the fraction fit into the number?* This is equivalent to **multiplying the number by the reciprocal of the fraction**.

The **reciprocal** of a fraction is obtained by **flipping the numerator and the denominator**. So, the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

---

### Step 2: Applying the Rule

Given the expression:

$$
10 \div \frac{2}{3}
$$

We apply the rule:

$$
10 \div \frac{2}{3} = 10 \times \frac{3}{2}
$$

---

### Step 3: Performing the Multiplication

Now, compute the multiplication:

$$
10 \times \frac{3}{2} = \frac{10 \times 3}{2} = \frac{30}{2} = 15
$$

So, the result of dividing $10$ by $\frac{2}{3}$ is $15$.

---

### Final Answer

To get the answer, Remmy should multiply $10$ by the reciprocal of $\frac{2}{3}$, which is:

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

Solution: Remember that dividing by a fraction is the same thing as multiplying by the reciprocal of the fraction.  The reciprocal of $\frac{2}{3}$ is $\boxed{\frac{3}{2}}$, so that is what Remmy should multiply by.