Explanation:
Okay, so I need to find the remainder when (n + 2)(n + 4)(n + 6) is divided by 7, given that n is congruent to 2 modulo 7. Let me think about how to approach this step by step.

First, since we're dealing with modular arithmetic, maybe I can substitute n with its equivalent value modulo 7. The problem says n ≡ 2 mod 7, which means that n can be written as 7k + 2 for some integer k. But maybe instead of working with n directly, I can replace n with 2 in the expression because when you take modulo 7, the value of n is equivalent to 2. Let me check if that's valid.

Yes, in modular arithmetic, if n ≡ 2 mod 7, then any expression involving n can be simplified by replacing n with 2. So, substituting n = 2 into the expression (n + 2)(n + 4)(n + 6) should give me the same result modulo 7. Let me compute that.

First, compute each term:

n + 2 = 2 + 2 = 4

n + 4 = 2 + 4 = 6

n + 6 = 2 + 6 = 8

So the product becomes 4 * 6 * 8. Now compute this product step by step.

First, 4 * 6 = 24. Then, 24 * 8 = 192. Now, find 192 mod 7. Hmm, how do I compute that?

Well, 7*27 = 189. So 192 - 189 = 3. Therefore, 192 mod 7 is 3. So the remainder is 3? Wait, but let me verify this because sometimes when dealing with modular arithmetic, there might be a trick or something I missed.

Alternatively, maybe I can compute each term modulo 7 first before multiplying. Let me try that approach again to confirm.

Original expression: (n + 2)(n + 4)(n + 6) mod 7.

Since n ≡ 2 mod 7, then:

n + 2 ≡ 2 + 2 = 4 mod 7.

n + 4 ≡ 2 + 4 = 6 mod 7.

n + 6 ≡ 2 + 6 = 8 ≡ 1 mod 7 (since 8 - 7 = 1). Wait, that's different from earlier. Wait, 8 mod 7 is 1? Wait, 7 goes into 8 once with remainder 1. Yes. So n + 6 ≡ 1 mod 7. Then the product is 4 * 6 * 1 mod 7.

So that's 4 * 6 = 24, and then 24 * 1 = 24. Then 24 mod 7 is... 7*3=21, 24-21=3. So again, 3. So that's the same answer as before. So maybe the remainder is 3? But let me check again, maybe I made a mistake in substitution.

Alternatively, perhaps there's a different approach. Let me think.

Since n ≡ 2 mod 7, let me consider the three terms:

(n + 2), (n + 4), (n + 6). These are three consecutive even numbers? Wait, not necessarily. Let me see. If n is congruent to 2 mod 7, then adding 2, 4, 6 gives numbers that are 2 + 2, 2 + 4, 2 + 6. So, they are 4, 6, 8. Which are 4, 6, and 8. But in mod 7, 8 is 1, so 4, 6, 1. Then multiplying them together. So 4*6*1=24≡3 mod7. So that's the same result.

Alternatively, maybe there's another way to think about it. Let me think about expanding the product algebraically. Let's try that.

Let me expand (n + 2)(n + 4)(n + 6). First, multiply the first two terms:

(n + 2)(n + 4) = n^2 + 6n + 8. Then multiply by (n + 6):

(n^2 + 6n + 8)(n + 6) = n^3 + 6n^2 + 8n + 6n^2 + 36n + 48 = n^3 + 12n^2 + 44n + 48.

But perhaps this is more complicated. However, since we are working modulo 7, maybe we can reduce coefficients mod 7.

Let me try that. Let me write the polynomial:

n^3 + 12n^2 + 44n + 48.

Now reduce each coefficient modulo 7.

First, 12 mod 7: 7*1=7, 12-7=5. So 12 ≡ 5 mod7.

44 mod7: 7*6=42, 44-42=2. So 44 ≡2 mod7.

48 mod7: 7*6=42, 48-42=6. So 48≡6 mod7.

So the polynomial becomes:

n^3 + 5n^2 + 2n + 6 mod7.

Now, since n ≡2 mod7, substitute n=2 into this polynomial:

2^3 + 5*(2^2) + 2*(2) + 6.

Compute each term:

2^3=8

5*(2^2)=5*4=20

2*(2)=4

So adding them up: 8 + 20 + 4 + 6 = 38.

Now, compute 38 mod7. 7*5=35, 38-35=3. So again, 3. Same answer. So that seems consistent.

Alternatively, maybe there's a smarter way using properties of modular arithmetic. Let me think. Since n ≡2 mod7, let's note that n + 2 ≡4 mod7, n +4≡6 mod7, n +6≡1 mod7 as before. Then 4*6*1=24≡3 mod7.

Alternatively, maybe we can think in terms of negative numbers? For example, in mod7, 6 is equivalent to -1, 4 is equivalent to -3, and 1 is equivalent to 1. Wait, so maybe 4*6*1 can be written as (-3)*(-1)*1=3*1=3 mod7. Which is same as before.

So this seems to confirm again that the remainder is 3. So maybe I should check with actual numbers. Let me pick a number n that is congruent to 2 mod7 and compute the product.

For example, take n=2. Then (2 + 2)(2 + 4)(2 +6)=4*6*8=192. Divide by 7, 7*27=189, 192-189=3. So remainder is 3. Correct.

Another example: n=9 (since 9≡2 mod7, since 9-2=7). Then compute (9 + 2)(9 +4)(9 +6)=11*13*15. Let me compute this:

First, 11 mod7 is 4, 13 mod7 is 6, 15 mod7 is 1. Then 4*6*1=24≡3 mod7. Alternatively, compute actual product:

11*13=143, 143*15. Let me compute 143*15. Well, 143*10=1430, 143*5=715, total is 1430+715=2145. Now divide 2145 by 7. Let me see: 7*300=2100. 2145-2100=45. 45 divided by 7 is 6*7=42, remainder 3. So total remainder is 3. Correct again.

Another example: take n= -5, since -5 ≡2 mod7? Wait, -5 +7=2, yes. So n=-5. Then compute (-5 +2)(-5 +4)(-5 +6)= (-3)(-1)(1)= (-3)*(-1)=3, then 3*1=3. Which is 3 mod7. So again, same result. So regardless of the value of n, as long as n≡2 mod7, the product is 3 mod7.

Therefore, I think that the remainder is 3. So the answer is 3. Wait, but let me check if there's another approach.

Alternatively, note that (n + 2)(n + 4)(n + 6) can be considered as the product of three consecutive numbers with a difference of 2. But since we are working mod7, maybe there's some pattern.

But since n ≡2 mod7, then n=2, so adding 2,4,6 gives numbers 4,6,8≡1 mod7. So the product is 4*6*1=24≡3. Alternatively, if I think about the numbers 4,6,1 mod7. Alternatively, 4*6=24≡3, then 3*1=3. So same result.

Alternatively, maybe there's a formula or identity that could be used here. For example, if we consider that (n + 2)(n + 4)(n + 6) is similar to expanding something. But maybe not necessary.

Alternatively, since we are dealing with mod7, maybe we can think about the multiplicative inverses or something. However, since the numbers 4, 6, 1 are all invertible modulo7 (since they are not multiples of 7), but I don't know if that helps here.

Alternatively, maybe notice that (n + 2)(n + 4)(n + 6) is similar to (n + 2)(n + 4)(n + 6) = (n + 2)(n + 4)(n + 6). If we set m = n + 2, then m ≡4 mod7. Then the product is m*(m + 2)*(m + 4). But not sure if that helps.

Alternatively, since m ≡4 mod7, m + 2 ≡6 mod7, m +4≡1 mod7, so again product is 4*6*1=24≡3 mod7. Same result.

Alternatively, if I think about factorials or something. But maybe overcomplicating.

Alternatively, think that since n ≡2 mod7, then n is even or odd? Well, 2 is even, so n is even. But not sure if that's relevant here.

Alternatively, think of the product as (n + 2)(n + 4)(n + 6) mod7. Since n ≡2, substitute:

(2 + 2)(2 + 4)(2 + 6) mod7 = 4*6*8 mod7. As before. Which is same as 4*6*1=24≡3.

So all methods point towards 3. Therefore, I think the answer is 3. Wait, but let me think if there's a chance that I made an error in calculation. For example, 4*6 is 24, then 24*1 is 24. 24 divided by 7 is 3*7=21, remainder 3. Yes, that's correct. 24 mod7=3. So yes.

Alternatively, is there a different interpretation of the problem? The problem says "the remainder when (n + 2)(n + 4)(n + 6) is divided by 7". So yes, it's straightforward. So given that n≡2 mod7, substitute into the expression and compute mod7. So answer is 3.

Alternatively, if I think of the general case. Suppose n ≡a mod7, then find (n + 2)(n + 4)(n + 6) mod7. But in this case, a=2. However, maybe the problem is designed such that regardless of a, there is some pattern? Wait, maybe not. But in this specific case, with a=2, the answer is 3.

Alternatively, maybe there's a formula for the product of three numbers spaced by 2 apart. But again, not sure. But in this specific case, with n ≡2 mod7, we can compute directly.

Therefore, after multiple methods and checks, I think the remainder is 3. Therefore, the answer is 3.

**Final Answer**
The remainder is \boxed{3}.

To solve the problem, we are given that $ n \equiv 2 \pmod{7} $, and we need to find the remainder when $ (n + 2)(n + 4)(n + 6) $ is divided by 7.

---

### Step 1: Leverage Modular Equivalence

Since $ n \equiv 2 \pmod{7} $, we can substitute $ n = 2 $ (in the context of modulo 7) into the expression. This is valid because we're only interested in the value of the expression modulo 7, and modular equivalence allows us to simplify components of the expression accordingly.

Thus, we evaluate:
$$
(n + 2)(n + 4)(n + 6) \equiv (2 + 2)(2 + 4)(2 + 6) \pmod{7}
$$

---

### Step 2: Simplify the Individual Terms Modulo 7

Now compute each term:

- $ n + 2 = 2 + 2 = 4 $
- $ n + 4 = 2 + 4 = 6 $
- $ n + 6 = 2 + 6 = 8 \equiv 1 \pmod{7} $ (since $ 8 - 7 = 1 $)

So the expression becomes:
$$
(4)(6)(1) \pmod{7}
$$

---

### Step 3: Multiply the Simplified Values

Now compute the product modulo 7:

$$
4 \times 6 = 24 \quad \text{and} \quad 24 \times 1 = 24
$$

So, we now find:
$$
24 \pmod{7}
$$

Since $ 7 \times 3 = 21 $, and $ 24 - 21 = 3 $, we have:
$$
24 \equiv 3 \pmod{7}
$$

---

### Final Answer

Therefore, the remainder when $ (n + 2)(n + 4)(n + 6) $ is divided by 7 is:

$$
\boxed{3}
$$

[Assistant]
Explanation:
Okay, so I need to find the remainder when (n + 2)(n + 4)(n + 6) is divided by 7, given that n is congruent to 2 modulo 7. Let me think about how to approach this step by step.

First, since we're dealing with modular arithmetic, maybe I can substitute n with its equivalent value modulo 7. The problem says n ≡ 2 mod 7, which means that n can be written as 7k + 2 for some integer k. But maybe instead of working with n directly, I can replace n with 2 in the expression because when you take modulo 7, the value of n is equivalent to 2. Let me check if that's valid.

Yes, in modular arithmetic, if n ≡ 2 mod 7, then any expression involving n can be simplified by replacing n with 2. So, substituting n = 2 into the expression (n + 2)(n + 4)(n + 6) should give me the same result modulo 7. Let me compute that.

First, compute each term:

n + 2 = 2 + 2 = 4

n + 4 = 2 + 4 = 6

n + 6 = 2 + 6 = 8

So the product becomes 4 * 6 * 8. Now compute this product step by step.

First, 4 * 6 = 24. Then, 24 * 8 = 192. Now, find 192 mod 7. Hmm, how do I compute that?

Well, 7*27 = 189. So 192 - 189 = 3. Therefore, 192 mod 7 is 3. So the remainder is 3? Wait, but let me verify this because sometimes when dealing with modular arithmetic, there might be a trick or something I missed.

Alternatively, maybe I can compute each term modulo 7 first before multiplying. Let me try that approach again to confirm.

Original expression: (n + 2)(n + 4)(n + 6) mod 7.

Since n ≡ 2 mod 7, then:

n + 2 ≡ 2 + 2 = 4 mod 7.

n + 4 ≡ 2 + 4 = 6 mod 7.

n + 6 ≡ 2 + 6 = 8 ≡ 1 mod 7 (since 8 - 7 = 1). Wait, that's different from earlier. Wait, 8 mod 7 is 1? Wait, 7 goes into 8 once with remainder 1. Yes. So n + 6 ≡ 1 mod 7. Then the product is 4 * 6 * 1 mod 7.

So that's 4 * 6 = 24, and then 24 * 1 = 24. Then 24 mod 7 is... 7*3=21, 24-21=3. So again, 3. So that's the same answer as before. So maybe the remainder is 3? But let me check again, maybe I made a mistake in substitution.

Alternatively, perhaps there's a different approach. Let me think.

Since n ≡ 2 mod 7, let me consider the three terms:

(n + 2), (n + 4), (n + 6). These are three consecutive even numbers? Wait, not necessarily. Let me see. If n is congruent to 2 mod 7, then adding 2, 4, 6 gives numbers that are 2 + 2, 2 + 4, 2 + 6. So, they are 4, 6, 8. Which are 4, 6, and 8. But in mod 7, 8 is 1, so 4, 6, 1. Then multiplying them together. So 4*6*1=24≡3 mod7. So that's the same result.

Alternatively, maybe there's another way to think about it. Let me think about expanding the product algebraically. Let's try that.

Let me expand (n + 2)(n + 4)(n + 6). First, multiply the first two terms:

(n + 2)(n + 4) = n^2 + 6n + 8. Then multiply by (n + 6):

(n^2 + 6n + 8)(n + 6) = n^3 + 6n^2 + 8n + 6n^2 + 36n + 48 = n^3 + 12n^2 + 44n + 48.

But perhaps this is more complicated. However, since we are working modulo 7, maybe we can reduce coefficients mod 7.

Let me try that. Let me write the polynomial:

n^3 + 12n^2 + 44n + 48.

Now reduce each coefficient modulo 7.

First, 12 mod 7: 7*1=7, 12-7=5. So 12 ≡ 5 mod7.

44 mod7: 7*6=42, 44-42=2. So 44 ≡2 mod7.

48 mod7: 7*6=42, 48-42=6. So 48≡6 mod7.

So the polynomial becomes:

n^3 + 5n^2 + 2n + 6 mod7.

Now, since n ≡2 mod7, substitute n=2 into this polynomial:

2^3 + 5*(2^2) + 2*(2) + 6.

Compute each term:

2^3=8

5*(2^2)=5*4=20

2*(2)=4

So adding them up: 8 + 20 + 4 + 6 = 38.

Now, compute 38 mod7. 7*5=35, 38-35=3. So again, 3. Same answer. So that seems consistent.

Alternatively, maybe there's a smarter way using properties of modular arithmetic. Let me think. Since n ≡2 mod7, let's note that n + 2 ≡4 mod7, n +4≡6 mod7, n +6≡1 mod7 as before. Then 4*6*1=24≡3 mod7.

Alternatively, maybe we can think in terms of negative numbers? For example, in mod7, 6 is equivalent to -1, 4 is equivalent to -3, and 1 is equivalent to 1. Wait, so maybe 4*6*1 can be written as (-3)*(-1)*1=3*1=3 mod7. Which is same as before.

So this seems to confirm again that the remainder is 3. So maybe I should check with actual numbers. Let me pick a number n that is congruent to 2 mod7 and compute the product.

For example, take n=2. Then (2 + 2)(2 + 4)(2 +6)=4*6*8=192. Divide by 7, 7*27=189, 192-189=3. So remainder is 3. Correct.

Another example: n=9 (since 9≡2 mod7, since 9-2=7). Then compute (9 + 2)(9 +4)(9 +6)=11*13*15. Let me compute this:

First, 11 mod7 is 4, 13 mod7 is 6, 15 mod7 is 1. Then 4*6*1=24≡3 mod7. Alternatively, compute actual product:

11*13=143, 143*15. Let me compute 143*15. Well, 143*10=1430, 143*5=715, total is 1430+715=2145. Now divide 2145 by 7. Let me see: 7*300=2100. 2145-2100=45. 45 divided by 7 is 6*7=42, remainder 3. So total remainder is 3. Correct again.

Another example: take n= -5, since -5 ≡2 mod7? Wait, -5 +7=2, yes. So n=-5. Then compute (-5 +2)(-5 +4)(-5 +6)= (-3)(-1)(1)= (-3)*(-1)=3, then 3*1=3. Which is 3 mod7. So again, same result. So regardless of the value of n, as long as n≡2 mod7, the product is 3 mod7.

Therefore, I think that the remainder is 3. So the answer is 3. Wait, but let me check if there's another approach.

Alternatively, note that (n + 2)(n + 4)(n + 6) can be considered as the product of three consecutive numbers with a difference of 2. But since we are working mod7, maybe there's some pattern.

But since n ≡2 mod7, then n=2, so adding 2,4,6 gives numbers 4,6,8≡1 mod7. So the product is 4*6*1=24≡3. Alternatively, if I think about the numbers 4,6,1 mod7. Alternatively, 4*6=24≡3, then 3*1=3. So same result.

Alternatively, maybe there's a formula or identity that could be used here. For example, if we consider that (n + 2)(n + 4)(n + 6) is similar to expanding something. But maybe not necessary.

Alternatively, since we are dealing with mod7, maybe we can think about the multiplicative inverses or something. However, since the numbers 4, 6, 1 are all invertible modulo7 (since they are not multiples of 7), but I don't know if that helps here.

Alternatively, maybe notice that (n + 2)(n + 4)(n + 6) is similar to (n + 2)(n + 4)(n + 6) = (n + 2)(n + 4)(n + 6). If we set m = n + 2, then m ≡4 mod7. Then the product is m*(m + 2)*(m + 4). But not sure if that helps.

Alternatively, since m ≡4 mod7, m + 2 ≡6 mod7, m +4≡1 mod7, so again product is 4*6*1=24≡3 mod7. Same result.

Alternatively, if I think about factorials or something. But maybe overcomplicating.

Alternatively, think that since n ≡2 mod7, then n is even or odd? Well, 2 is even, so n is even. But not sure if that's relevant here.

Alternatively, think of the product as (n + 2)(n + 4)(n + 6) mod7. Since n ≡2, substitute:

(2 + 2)(2 + 4)(2 + 6) mod7 = 4*6*8 mod7. As before. Which is same as 4*6*1=24≡3.

So all methods point towards 3. Therefore, I think the answer is 3. Wait, but let me think if there's a chance that I made an error in calculation. For example, 4*6 is 24, then 24*1 is 24. 24 divided by 7 is 3*7=21, remainder 3. Yes, that's correct. 24 mod7=3. So yes.

Alternatively, is there a different interpretation of the problem? The problem says "the remainder when (n + 2)(n + 4)(n + 6) is divided by 7". So yes, it's straightforward. So given that n≡2 mod7, substitute into the expression and compute mod7. So answer is 3.

Alternatively, if I think of the general case. Suppose n ≡a mod7, then find (n + 2)(n + 4)(n + 6) mod7. But in this case, a=2. However, maybe the problem is designed such that regardless of a, there is some pattern? Wait, maybe not. But in this specific case, with a=2, the answer is 3.

Alternatively, maybe there's a formula for the product of three numbers spaced by 2 apart. But again, not sure. But in this specific case, with n ≡2 mod7, we can compute directly.

Therefore, after multiple methods and checks, I think the remainder is 3. Therefore, the answer is 3.

**Final Answer**
The remainder is \boxed{3}.

To solve the problem, we are given that $ n \equiv 2 \pmod{7} $, and we need to find the remainder when $ (n + 2)(n + 4)(n + 6) $ is divided by 7.

---

### Step 1: Leverage Modular Equivalence

Since $ n \equiv 2 \pmod{7} $, we can substitute $ n = 2 $ (in the context of modulo 7) into the expression. This is valid because we're only interested in the value of the expression modulo 7, and modular equivalence allows us to simplify components of the expression accordingly.

Thus, we evaluate:
$$
(n + 2)(n + 4)(n + 6) \equiv (2 + 2)(2 + 4)(2 + 6) \pmod{7}
$$

---

### Step 2: Simplify the Individual Terms Modulo 7

Now compute each term:

- $ n + 2 = 2 + 2 = 4 $
- $ n + 4 = 2 + 4 = 6 $
- $ n + 6 = 2 + 6 = 8 \equiv 1 \pmod{7} $ (since $ 8 - 7 = 1 $)

So the expression becomes:
$$
(4)(6)(1) \pmod{7}
$$

---

### Step 3: Multiply the Simplified Values

Now compute the product modulo 7:

$$
4 \times 6 = 24 \quad \text{and} \quad 24 \times 1 = 24
$$

So, we now find:
$$
24 \pmod{7}
$$

Since $ 7 \times 3 = 21 $, and $ 24 - 21 = 3 $, we have:
$$
24 \equiv 3 \pmod{7}
$$

---

### Final Answer

Therefore, the remainder when $ (n + 2)(n + 4)(n + 6) $ is divided by 7 is:

$$
\boxed{3}
$$

Solution: If $n \equiv 2 \pmod{7}$, then $(n + 2)(n + 4)(n + 6) \equiv 4 \cdot 6 \cdot 8 \equiv 4 \cdot 6 \cdot 1 \equiv 24 \equiv \boxed{3} \pmod{7}$.


Explanation:
Okay, so I need to find the roots of the equation $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$. Hmm, let me think about how to approach this.

First, maybe I should expand all the terms to simplify the equation. Let me recall the formula for expanding cubes. The cube of a binomial is $ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $. But since these are differences, maybe I can use that formula as well. Let me try expanding each term step by step.

Starting with the left side: $(x - 3)^3 + (x - 7)^3$. Let me compute each cube separately.

First, expand $(x - 3)^3$:

$$
(x - 3)^3 = x^3 - 3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2 - 3^3
$$
Wait, let me check that again. The formula is $ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $. So substituting $ a = x $, $ b = 3 $:

$$
(x - 3)^3 = x^3 - 3x^2 \cdot 3 + 3x \cdot 3^2 - 3^3 = x^3 - 9x^2 + 27x - 27
$$

Similarly, expanding $(x - 7)^3$:

$$
(x - 7)^3 = x^3 - 3x^2 \cdot 7 + 3x \cdot 7^2 - 7^3 = x^3 - 21x^2 + 147x - 343
$$

So adding these two together:

Left side becomes:

$$
(x^3 - 9x^2 + 27x - 27) + (x^3 - 21x^2 + 147x - 343) = 2x^3 - 30x^2 + 174x - 370
$$

Okay, now the right side is $(2x - 10)^3$. Let me expand that. Again, using the cube formula:

$$
(2x - 10)^3 = (2x)^3 - 3(2x)^2 \cdot 10 + 3(2x) \cdot 10^2 - 10^3
$$

Wait, hold on. Let me be careful. The formula is $ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $. Here, $ a = 2x $, $ b = 10 $:

So:

$$
(2x - 10)^3 = (2x)^3 - 3(2x)^2 \cdot 10 + 3(2x)(10)^2 - (10)^3
$$
Calculating each term:

First term: $ (2x)^3 = 8x^3 $

Second term: $ 3(2x)^2 \cdot 10 = 3 \cdot 4x^2 \cdot 10 = 120x^2 $, but since it's minus, it's $ -120x^2 $

Third term: $ 3(2x)(10)^2 = 3 \cdot 2x \cdot 100 = 600x $

Fourth term: $ -10^3 = -1000 $

So altogether, $(2x - 10)^3 = 8x^3 - 120x^2 + 600x - 1000$

Now, our original equation is:

Left side: $ 2x^3 - 30x^2 + 174x - 370 $

Right side: $ 8x^3 - 120x^2 + 600x - 1000 $

So moving everything to one side:

Left side minus right side equals zero:

$$
(2x^3 - 30x^2 + 174x - 370) - (8x^3 - 120x^2 + 600x - 1000) = 0
$$

Let me compute each term:

2x^3 - 8x^3 = -6x^3

-30x^2 - (-120x^2) = -30x^2 + 120x^2 = 90x^2

174x - 600x = -426x

-370 - (-1000) = -370 + 1000 = 630

So putting it all together:

-6x^3 + 90x^2 - 426x + 630 = 0

Hmm, that's a cubic equation. Let me write it as:

-6x^3 + 90x^2 - 426x + 630 = 0

It might be easier to multiply both sides by -1 to make the leading coefficient positive:

6x^3 - 90x^2 + 426x - 630 = 0

Now, perhaps we can factor out a common factor. Let me check if all coefficients are divisible by something. Let's see:

6, 90, 426, 630.

Divide by 6:

6x^3 / 6 = x^3

-90x^2 / 6 = -15x^2

426x / 6 = 71x

-630 / 6 = -105

So equation becomes:

x^3 - 15x^2 + 71x - 105 = 0

Hmm, so now we have a cubic equation: $ x^3 - 15x^2 + 71x - 105 = 0 $

I need to find the roots of this cubic. Maybe there are rational roots. Let me check using the Rational Root Theorem. The possible rational roots are factors of 105 divided by factors of 1 (the leading coefficient). So possible roots are ±1, ±3, ±5, ±7, ±15, ±21, ±35, ±105.

Let me test them one by one.

First, x = 1:

1 - 15 + 71 - 105 = 1 -15 is -14, +71 is 57, -105 is -48 ≠ 0

x = 3:

27 - 15*(9) + 71*3 - 105 = 27 - 135 + 213 - 105

Compute step by step:

27 - 135 = -108

-108 + 213 = 105

105 - 105 = 0

Oh, x=3 is a root!

So (x - 3) is a factor. Let's perform polynomial division or use synthetic division to factor it out.

Let me use synthetic division with root 3.

Coefficients of the polynomial x^3 -15x^2 +71x -105 are 1, -15, 71, -105

Set up synthetic division:

3 | 1   -15   71   -105
  |     3   -36   105
  -----------------------
    1   -12   35     0

So the quotient polynomial is x^2 -12x +35.

Therefore, the cubic factors as (x - 3)(x^2 -12x +35) = 0

Now, set each factor equal to zero:

x - 3 = 0 => x = 3

x^2 -12x +35 = 0

Solve the quadratic equation:

x = [12 ± sqrt(144 - 140)] / 2 = [12 ± sqrt(4)] / 2 = [12 ± 2]/2 = 14/2=7 or 10/2=5

So the roots are x = 3, x = 5, x = 7.

Wait, but let me check if these are all valid roots. Because sometimes when we manipulate equations, especially when expanding, we might introduce extraneous roots? Wait, but here we just expanded both sides and moved everything to one side, so the roots should all be valid. However, let me check each root in the original equation to make sure.

First, check x=3:

Left side: (3 - 3)^3 + (3 - 7)^3 = 0 + (-4)^3 = -64

Right side: (2*3 -10)^3 = (6 -10)^3 = (-4)^3 = -64. So  -64 = -64, which is okay.

Check x=5:

Left side: (5 - 3)^3 + (5 -7)^3 = (2)^3 + (-2)^3 = 8 + (-8) = 0

Right side: (2*5 -10)^3 = (10 -10)^3 = 0^3 = 0. So 0=0, good.

Check x=7:

Left side: (7 -3)^3 + (7 -7)^3 = 4^3 + 0 = 64 + 0 = 64

Right side: (2*7 -10)^3 = (14 -10)^3 = 4^3 = 64. 64=64, correct.

So all three roots are valid. So the equation has three real roots: 3, 5, 7.

But let me think again. Wait, the original equation was $(x - 3)^3 + (x -7)^3 = (2x -10)^3$. Let me check if there is some symmetry here.

Notice that 2x -10 = 2(x - 5). So maybe there's some relationship with x=5? Also, the roots are 3,5,7. Which are symmetric around 5? 3 and 7 are equidistant from 5. So that might be a clue. Let me check if there's a way to see this without expanding.

Alternatively, perhaps we can use the identity for $ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $, but here we have $ a^3 + b^3 = c^3 $. Maybe not directly helpful. Alternatively, maybe substitution.

Let me let y = x - 5, shifting the variable to make it symmetric. Let me try that.

Let x = y + 5. Then:

Left side:

(x - 3)^3 + (x -7)^3 = (y + 5 -3)^3 + (y +5 -7)^3 = (y + 2)^3 + (y - 2)^3

Right side:

(2x - 10)^3 = 2(x - 5)^3 = 2y^3

So equation becomes:

(y + 2)^3 + (y - 2)^3 = 2y^3

Compute left side:

Expand (y + 2)^3 = y^3 + 6y^2 + 12y + 8

(y - 2)^3 = y^3 - 6y^2 + 12y -8

Add them together:

y^3 +6y^2 +12y +8 + y^3 -6y^2 +12y -8 = 2y^3 + 24y

Thus, equation becomes:

2y^3 +24y = 2y^3

Subtract 2y^3 from both sides:

24y = 0 => y = 0

So y=0 => x = 5. Wait, but earlier we found three roots. What's going on here?

Wait, this suggests that only y=0 is a solution? But earlier expansion gave three solutions. There must be an error here. Wait, no. Wait, when I did substitution, maybe I made a mistake? Let me check again.

Original equation:

(y + 2)^3 + (y - 2)^3 = 2y^3

Left side expansion:

(y + 2)^3 + (y - 2)^3 = [y^3 + 3y^2*2 + 3y*4 + 8] + [y^3 - 3y^2*2 + 3y*4 - 8] ?

Wait, wait, no. Wait, let me do it step by step again.

Wait, (y + 2)^3 is y^3 + 3y²(2) + 3y(2²) + 2³ = y³ + 6y² + 12y + 8

Similarly, (y - 2)^3 = y³ - 6y² + 12y -8

Adding them: y³ +6y² +12y +8 + y³ -6y² +12y -8 = 2y³ + 24y. That's correct. So equation is 2y³ +24y = 2y³ ⇒ 24y = 0 ⇒ y = 0. But according to previous expansion, there are three roots. How come?

Wait, this seems contradictory. Because according to substitution approach, only y=0 is the solution, but earlier expansion showed x=3,5,7. Which is y=-2, 0, +2. So why does substitution method only show y=0? Maybe because of an error in substitution?

Wait, let me check with original equation. If x=3, then y = 3 -5 = -2. Let me plug y=-2 into equation:

Left side: (-2 + 2)^3 + (-2 -2)^3 = (0)^3 + (-4)^3 = 0 + (-64) = -64

Right side: 2*(-2)^3 = 2*(-8) = -16. Wait, but that's not equal. Wait, hold on, that can't be. Wait, according to substitution, we had equation 2y³ +24y = 2y³ ⇒ 24y = 0. But if we plug y=-2 into original transformed equation: (y + 2)^3 + (y - 2)^3 should equal 2y³?

Wait, let me check with y=-2:

Left side: (-2 +2)^3 + (-2 -2)^3 = 0^3 + (-4)^3 = 0 -64 = -64

Right side: 2*(-2)^3 = 2*(-8) = -16. So -64 ≠ -16. Which suggests something is wrong. But earlier, when we plugged x=3 into original equation, it worked. So there's inconsistency here. What's happening?

Wait, there must be an error in substitution. Let me check again. Original equation:

Left side: (x - 3)^3 + (x - 7)^3

Right side: (2x -10)^3.

We let x = y + 5. Then 2x -10 = 2(y +5) -10 = 2y +10 -10 = 2y. So right side is (2y)^3 = 8y³. Wait, earlier I thought it was 2y³. Wait, that's my mistake! Earlier I said 2(x -5)^3, which is 2y³, but actually, 2x -10 = 2(x - 5), so (2x -10)^3 = [2(x - 5)]³ = 8(x -5)^3 = 8y³. So the right side is 8y³, not 2y³.

Oh! There was a mistake here. I incorrectly said right side is 2y³, but it's actually 8y³. So that explains the discrepancy. Let me redo the substitution approach with the correct right side.

So transformed equation:

Left side: (y + 2)^3 + (y - 2)^3 = 2y³ +24y (as before)

Right side: 8y³

Therefore equation is:

2y³ +24y = 8y³ ⇒ 0 = 6y³ -24y ⇒ 6y³ -24y = 0 ⇒ 6y(y² -4) = 0 ⇒ 6y(y - 2)(y + 2) = 0

Thus, solutions are y = 0, y = 2, y = -2. Therefore, x = y +5:

y=0 ⇒ x=5

y=2 ⇒ x=7

y=-2 ⇒ x=3

Which matches the previous results! So that's why substitution gives us three roots. Initially, I had made a mistake in calculating the right-hand side, which led to confusion. So this substitution method is more straightforward and shows that the roots are x=3,5,7.

So, either way, the roots are 3,5,7. Therefore, the answer is these three numbers.

But just to make sure, let me verify again with substitution approach. For example, take y=2:

Left side: (2 +2)^3 + (2 -2)^3 = 4³ +0 = 64

Right side: 8*(2)^3 = 8*8 = 64. Correct.

Similarly for y=-2:

Left side: (-2 + 2)^3 + (-2 -2)^3 = 0 + (-4)^3 = -64

Right side: 8*(-2)^3 = 8*(-8) = -64. Correct.

So that's consistent. So substitution approach confirms the roots. So that's a better method.

But since I first expanded everything, got the cubic, factored it, found roots, and checked them all, and substitution gives same result, so I can be confident that the roots are indeed 3,5,7.

Therefore, the roots of the equation are 3, 5, and 7.

**Final Answer**
The roots are $\boxed{3}$, $\boxed{5}$, and $\boxed{7}$.

To solve the equation:

$$
(x - 3)^3 + (x - 7)^3 = (2x - 10)^3
$$

we aim to find all real roots $ x $ that satisfy this cubic equation. We begin by simplifying the expressions on both sides.

---

### Step 1: Expand both sides

#### Left Side:

- $(x - 3)^3 = x^3 - 9x^2 + 27x - 27$
- $(x - 7)^3 = x^3 - 21x^2 + 147x - 343$

So the left-hand side becomes:

$$
(x - 3)^3 + (x - 7)^3 = 2x^3 - 30x^2 + 174x - 370
$$

#### Right Side:

- $(2x - 10)^3 = (2x)^3 - 3(2x)^2(10) + 3(2x)(10)^2 - (10)^3$
- Simplifies to:

$$
(2x - 10)^3 = 8x^3 - 120x^2 + 600x - 1000
$$

---

### Step 2: Move all terms to one side

Subtracting the right side from the left:

$$
(2x^3 - 30x^2 + 174x - 370) - (8x^3 - 120x^2 + 600x - 1000) = 0
$$

Compute term by term:

- $2x^3 - 8x^3 = -6x^3$
- $-30x^2 + 120x^2 = 90x^2$
- $174x - 600x = -426x$
- $-370 + 1000 = 630$

Thus, the equation becomes:

$$
-6x^3 + 90x^2 - 426x + 630 = 0
$$

Multiply both sides by $-1$ for standard form:

$$
6x^3 - 90x^2 + 426x - 630 = 0
$$

Divide the entire equation by 6 to simplify:

$$
x^3 - 15x^2 + 71x - 105 = 0
$$

---

### Step 3: Factor the cubic polynomial

Use the **Rational Root Theorem** to test possible rational roots. The possible rational roots are:

$$
\pm1, \pm3, \pm5, \pm7, \pm15, \pm21, \pm35, \pm105
$$

Try $ x = 3 $:

$$
3^3 - 15(3)^2 + 71(3) - 105 = 27 - 135 + 213 - 105 = 0
$$

So, $ x = 3 $ is a root. That means $ (x - 3) $ is a factor.

Use **synthetic division** to divide $ x^3 - 15x^2 + 71x - 105 $ by $ (x - 3) $:

$$
\begin{array}{r|rrrr}
3 & 1 & -15 & 71 & -105 \\
  &   & 3  & -36 & 105 \\
\hline
  & 1 & -12 & 35 & 0 \\
\end{array}
$$

So the cubic factors as:

$$
(x - 3)(x^2 - 12x + 35) = 0
$$

Now factor the quadratic:

$$
x^2 - 12x + 35 = (x - 5)(x - 7)
$$

So the full factorization is:

$$
(x - 3)(x - 5)(x - 7) = 0
$$

---

### Step 4: Identify the roots

Setting each factor equal to zero:

$$
x = 3,\quad x = 5,\quad x = 7
$$

---

### Step 5: Final Verification

Plug each value back into the original equation to confirm they are valid solutions.

- $ x = 3 $: $(3 - 3)^3 + (3 - 7)^3 = 0 + (-4)^3 = -64$, and $ (2\cdot3 - 10)^3 = -64 $
- $ x = 5 $: $(5 - 3)^3 + (5 - 7)^3 = 8 + (-8) = 0$, and $ (2\cdot5 -10)^3 = 0 $
- $ x = 7 $: $(7 - 3)^3 + (7 - 7)^3 = 64 + 0 = 64$, and $ (2\cdot7 -10)^3 = 64 $

All roots check out.

---

### Final Answer

$$
\boxed{3}, \boxed{5}, \boxed{7}
$$

[Assistant]
Explanation:
Okay, so I need to find the roots of the equation $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$. Hmm, let me think about how to approach this.

First, maybe I should expand all the terms to simplify the equation. Let me recall the formula for expanding cubes. The cube of a binomial is $ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $. But since these are differences, maybe I can use that formula as well. Let me try expanding each term step by step.

Starting with the left side: $(x - 3)^3 + (x - 7)^3$. Let me compute each cube separately.

First, expand $(x - 3)^3$:

$$
(x - 3)^3 = x^3 - 3 \cdot x^2 \cdot 3 + 3 \cdot x \cdot 3^2 - 3^3
$$
Wait, let me check that again. The formula is $ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $. So substituting $ a = x $, $ b = 3 $:

$$
(x - 3)^3 = x^3 - 3x^2 \cdot 3 + 3x \cdot 3^2 - 3^3 = x^3 - 9x^2 + 27x - 27
$$

Similarly, expanding $(x - 7)^3$:

$$
(x - 7)^3 = x^3 - 3x^2 \cdot 7 + 3x \cdot 7^2 - 7^3 = x^3 - 21x^2 + 147x - 343
$$

So adding these two together:

Left side becomes:

$$
(x^3 - 9x^2 + 27x - 27) + (x^3 - 21x^2 + 147x - 343) = 2x^3 - 30x^2 + 174x - 370
$$

Okay, now the right side is $(2x - 10)^3$. Let me expand that. Again, using the cube formula:

$$
(2x - 10)^3 = (2x)^3 - 3(2x)^2 \cdot 10 + 3(2x) \cdot 10^2 - 10^3
$$

Wait, hold on. Let me be careful. The formula is $ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $. Here, $ a = 2x $, $ b = 10 $:

So:

$$
(2x - 10)^3 = (2x)^3 - 3(2x)^2 \cdot 10 + 3(2x)(10)^2 - (10)^3
$$
Calculating each term:

First term: $ (2x)^3 = 8x^3 $

Second term: $ 3(2x)^2 \cdot 10 = 3 \cdot 4x^2 \cdot 10 = 120x^2 $, but since it's minus, it's $ -120x^2 $

Third term: $ 3(2x)(10)^2 = 3 \cdot 2x \cdot 100 = 600x $

Fourth term: $ -10^3 = -1000 $

So altogether, $(2x - 10)^3 = 8x^3 - 120x^2 + 600x - 1000$

Now, our original equation is:

Left side: $ 2x^3 - 30x^2 + 174x - 370 $

Right side: $ 8x^3 - 120x^2 + 600x - 1000 $

So moving everything to one side:

Left side minus right side equals zero:

$$
(2x^3 - 30x^2 + 174x - 370) - (8x^3 - 120x^2 + 600x - 1000) = 0
$$

Let me compute each term:

2x^3 - 8x^3 = -6x^3

-30x^2 - (-120x^2) = -30x^2 + 120x^2 = 90x^2

174x - 600x = -426x

-370 - (-1000) = -370 + 1000 = 630

So putting it all together:

-6x^3 + 90x^2 - 426x + 630 = 0

Hmm, that's a cubic equation. Let me write it as:

-6x^3 + 90x^2 - 426x + 630 = 0

It might be easier to multiply both sides by -1 to make the leading coefficient positive:

6x^3 - 90x^2 + 426x - 630 = 0

Now, perhaps we can factor out a common factor. Let me check if all coefficients are divisible by something. Let's see:

6, 90, 426, 630.

Divide by 6:

6x^3 / 6 = x^3

-90x^2 / 6 = -15x^2

426x / 6 = 71x

-630 / 6 = -105

So equation becomes:

x^3 - 15x^2 + 71x - 105 = 0

Hmm, so now we have a cubic equation: $ x^3 - 15x^2 + 71x - 105 = 0 $

I need to find the roots of this cubic. Maybe there are rational roots. Let me check using the Rational Root Theorem. The possible rational roots are factors of 105 divided by factors of 1 (the leading coefficient). So possible roots are ±1, ±3, ±5, ±7, ±15, ±21, ±35, ±105.

Let me test them one by one.

First, x = 1:

1 - 15 + 71 - 105 = 1 -15 is -14, +71 is 57, -105 is -48 ≠ 0

x = 3:

27 - 15*(9) + 71*3 - 105 = 27 - 135 + 213 - 105

Compute step by step:

27 - 135 = -108

-108 + 213 = 105

105 - 105 = 0

Oh, x=3 is a root!

So (x - 3) is a factor. Let's perform polynomial division or use synthetic division to factor it out.

Let me use synthetic division with root 3.

Coefficients of the polynomial x^3 -15x^2 +71x -105 are 1, -15, 71, -105

Set up synthetic division:

3 | 1   -15   71   -105
  |     3   -36   105
  -----------------------
    1   -12   35     0

So the quotient polynomial is x^2 -12x +35.

Therefore, the cubic factors as (x - 3)(x^2 -12x +35) = 0

Now, set each factor equal to zero:

x - 3 = 0 => x = 3

x^2 -12x +35 = 0

Solve the quadratic equation:

x = [12 ± sqrt(144 - 140)] / 2 = [12 ± sqrt(4)] / 2 = [12 ± 2]/2 = 14/2=7 or 10/2=5

So the roots are x = 3, x = 5, x = 7.

Wait, but let me check if these are all valid roots. Because sometimes when we manipulate equations, especially when expanding, we might introduce extraneous roots? Wait, but here we just expanded both sides and moved everything to one side, so the roots should all be valid. However, let me check each root in the original equation to make sure.

First, check x=3:

Left side: (3 - 3)^3 + (3 - 7)^3 = 0 + (-4)^3 = -64

Right side: (2*3 -10)^3 = (6 -10)^3 = (-4)^3 = -64. So  -64 = -64, which is okay.

Check x=5:

Left side: (5 - 3)^3 + (5 -7)^3 = (2)^3 + (-2)^3 = 8 + (-8) = 0

Right side: (2*5 -10)^3 = (10 -10)^3 = 0^3 = 0. So 0=0, good.

Check x=7:

Left side: (7 -3)^3 + (7 -7)^3 = 4^3 + 0 = 64 + 0 = 64

Right side: (2*7 -10)^3 = (14 -10)^3 = 4^3 = 64. 64=64, correct.

So all three roots are valid. So the equation has three real roots: 3, 5, 7.

But let me think again. Wait, the original equation was $(x - 3)^3 + (x -7)^3 = (2x -10)^3$. Let me check if there is some symmetry here.

Notice that 2x -10 = 2(x - 5). So maybe there's some relationship with x=5? Also, the roots are 3,5,7. Which are symmetric around 5? 3 and 7 are equidistant from 5. So that might be a clue. Let me check if there's a way to see this without expanding.

Alternatively, perhaps we can use the identity for $ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $, but here we have $ a^3 + b^3 = c^3 $. Maybe not directly helpful. Alternatively, maybe substitution.

Let me let y = x - 5, shifting the variable to make it symmetric. Let me try that.

Let x = y + 5. Then:

Left side:

(x - 3)^3 + (x -7)^3 = (y + 5 -3)^3 + (y +5 -7)^3 = (y + 2)^3 + (y - 2)^3

Right side:

(2x - 10)^3 = 2(x - 5)^3 = 2y^3

So equation becomes:

(y + 2)^3 + (y - 2)^3 = 2y^3

Compute left side:

Expand (y + 2)^3 = y^3 + 6y^2 + 12y + 8

(y - 2)^3 = y^3 - 6y^2 + 12y -8

Add them together:

y^3 +6y^2 +12y +8 + y^3 -6y^2 +12y -8 = 2y^3 + 24y

Thus, equation becomes:

2y^3 +24y = 2y^3

Subtract 2y^3 from both sides:

24y = 0 => y = 0

So y=0 => x = 5. Wait, but earlier we found three roots. What's going on here?

Wait, this suggests that only y=0 is a solution? But earlier expansion gave three solutions. There must be an error here. Wait, no. Wait, when I did substitution, maybe I made a mistake? Let me check again.

Original equation:

(y + 2)^3 + (y - 2)^3 = 2y^3

Left side expansion:

(y + 2)^3 + (y - 2)^3 = [y^3 + 3y^2*2 + 3y*4 + 8] + [y^3 - 3y^2*2 + 3y*4 - 8] ?

Wait, wait, no. Wait, let me do it step by step again.

Wait, (y + 2)^3 is y^3 + 3y²(2) + 3y(2²) + 2³ = y³ + 6y² + 12y + 8

Similarly, (y - 2)^3 = y³ - 6y² + 12y -8

Adding them: y³ +6y² +12y +8 + y³ -6y² +12y -8 = 2y³ + 24y. That's correct. So equation is 2y³ +24y = 2y³ ⇒ 24y = 0 ⇒ y = 0. But according to previous expansion, there are three roots. How come?

Wait, this seems contradictory. Because according to substitution approach, only y=0 is the solution, but earlier expansion showed x=3,5,7. Which is y=-2, 0, +2. So why does substitution method only show y=0? Maybe because of an error in substitution?

Wait, let me check with original equation. If x=3, then y = 3 -5 = -2. Let me plug y=-2 into equation:

Left side: (-2 + 2)^3 + (-2 -2)^3 = (0)^3 + (-4)^3 = 0 + (-64) = -64

Right side: 2*(-2)^3 = 2*(-8) = -16. Wait, but that's not equal. Wait, hold on, that can't be. Wait, according to substitution, we had equation 2y³ +24y = 2y³ ⇒ 24y = 0. But if we plug y=-2 into original transformed equation: (y + 2)^3 + (y - 2)^3 should equal 2y³?

Wait, let me check with y=-2:

Left side: (-2 +2)^3 + (-2 -2)^3 = 0^3 + (-4)^3 = 0 -64 = -64

Right side: 2*(-2)^3 = 2*(-8) = -16. So -64 ≠ -16. Which suggests something is wrong. But earlier, when we plugged x=3 into original equation, it worked. So there's inconsistency here. What's happening?

Wait, there must be an error in substitution. Let me check again. Original equation:

Left side: (x - 3)^3 + (x - 7)^3

Right side: (2x -10)^3.

We let x = y + 5. Then 2x -10 = 2(y +5) -10 = 2y +10 -10 = 2y. So right side is (2y)^3 = 8y³. Wait, earlier I thought it was 2y³. Wait, that's my mistake! Earlier I said 2(x -5)^3, which is 2y³, but actually, 2x -10 = 2(x - 5), so (2x -10)^3 = [2(x - 5)]³ = 8(x -5)^3 = 8y³. So the right side is 8y³, not 2y³.

Oh! There was a mistake here. I incorrectly said right side is 2y³, but it's actually 8y³. So that explains the discrepancy. Let me redo the substitution approach with the correct right side.

So transformed equation:

Left side: (y + 2)^3 + (y - 2)^3 = 2y³ +24y (as before)

Right side: 8y³

Therefore equation is:

2y³ +24y = 8y³ ⇒ 0 = 6y³ -24y ⇒ 6y³ -24y = 0 ⇒ 6y(y² -4) = 0 ⇒ 6y(y - 2)(y + 2) = 0

Thus, solutions are y = 0, y = 2, y = -2. Therefore, x = y +5:

y=0 ⇒ x=5

y=2 ⇒ x=7

y=-2 ⇒ x=3

Which matches the previous results! So that's why substitution gives us three roots. Initially, I had made a mistake in calculating the right-hand side, which led to confusion. So this substitution method is more straightforward and shows that the roots are x=3,5,7.

So, either way, the roots are 3,5,7. Therefore, the answer is these three numbers.

But just to make sure, let me verify again with substitution approach. For example, take y=2:

Left side: (2 +2)^3 + (2 -2)^3 = 4³ +0 = 64

Right side: 8*(2)^3 = 8*8 = 64. Correct.

Similarly for y=-2:

Left side: (-2 + 2)^3 + (-2 -2)^3 = 0 + (-4)^3 = -64

Right side: 8*(-2)^3 = 8*(-8) = -64. Correct.

So that's consistent. So substitution approach confirms the roots. So that's a better method.

But since I first expanded everything, got the cubic, factored it, found roots, and checked them all, and substitution gives same result, so I can be confident that the roots are indeed 3,5,7.

Therefore, the roots of the equation are 3, 5, and 7.

**Final Answer**
The roots are $\boxed{3}$, $\boxed{5}$, and $\boxed{7}$.

To solve the equation:

$$
(x - 3)^3 + (x - 7)^3 = (2x - 10)^3
$$

we aim to find all real roots $ x $ that satisfy this cubic equation. We begin by simplifying the expressions on both sides.

---

### Step 1: Expand both sides

#### Left Side:

- $(x - 3)^3 = x^3 - 9x^2 + 27x - 27$
- $(x - 7)^3 = x^3 - 21x^2 + 147x - 343$

So the left-hand side becomes:

$$
(x - 3)^3 + (x - 7)^3 = 2x^3 - 30x^2 + 174x - 370
$$

#### Right Side:

- $(2x - 10)^3 = (2x)^3 - 3(2x)^2(10) + 3(2x)(10)^2 - (10)^3$
- Simplifies to:

$$
(2x - 10)^3 = 8x^3 - 120x^2 + 600x - 1000
$$

---

### Step 2: Move all terms to one side

Subtracting the right side from the left:

$$
(2x^3 - 30x^2 + 174x - 370) - (8x^3 - 120x^2 + 600x - 1000) = 0
$$

Compute term by term:

- $2x^3 - 8x^3 = -6x^3$
- $-30x^2 + 120x^2 = 90x^2$
- $174x - 600x = -426x$
- $-370 + 1000 = 630$

Thus, the equation becomes:

$$
-6x^3 + 90x^2 - 426x + 630 = 0
$$

Multiply both sides by $-1$ for standard form:

$$
6x^3 - 90x^2 + 426x - 630 = 0
$$

Divide the entire equation by 6 to simplify:

$$
x^3 - 15x^2 + 71x - 105 = 0
$$

---

### Step 3: Factor the cubic polynomial

Use the **Rational Root Theorem** to test possible rational roots. The possible rational roots are:

$$
\pm1, \pm3, \pm5, \pm7, \pm15, \pm21, \pm35, \pm105
$$

Try $ x = 3 $:

$$
3^3 - 15(3)^2 + 71(3) - 105 = 27 - 135 + 213 - 105 = 0
$$

So, $ x = 3 $ is a root. That means $ (x - 3) $ is a factor.

Use **synthetic division** to divide $ x^3 - 15x^2 + 71x - 105 $ by $ (x - 3) $:

$$
\begin{array}{r|rrrr}
3 & 1 & -15 & 71 & -105 \\
  &   & 3  & -36 & 105 \\
\hline
  & 1 & -12 & 35 & 0 \\
\end{array}
$$

So the cubic factors as:

$$
(x - 3)(x^2 - 12x + 35) = 0
$$

Now factor the quadratic:

$$
x^2 - 12x + 35 = (x - 5)(x - 7)
$$

So the full factorization is:

$$
(x - 3)(x - 5)(x - 7) = 0
$$

---

### Step 4: Identify the roots

Setting each factor equal to zero:

$$
x = 3,\quad x = 5,\quad x = 7
$$

---

### Step 5: Final Verification

Plug each value back into the original equation to confirm they are valid solutions.

- $ x = 3 $: $(3 - 3)^3 + (3 - 7)^3 = 0 + (-4)^3 = -64$, and $ (2\cdot3 - 10)^3 = -64 $
- $ x = 5 $: $(5 - 3)^3 + (5 - 7)^3 = 8 + (-8) = 0$, and $ (2\cdot5 -10)^3 = 0 $
- $ x = 7 $: $(7 - 3)^3 + (7 - 7)^3 = 64 + 0 = 64$, and $ (2\cdot7 -10)^3 = 64 $

All roots check out.

---

### Final Answer

$$
\boxed{3}, \boxed{5}, \boxed{7}
$$

Solution:
Let $a = x - 3$ and $b = x - 7.$  Then we can write the given equation as
\[a^3 + b^3 = (a + b)^3.\]Expanding, we get $a^3 + b^3 = a^3 + 3a^2 b + 3ab^2 + b^3,$ so $3a^2 b + 3ab^2 = 0,$ which factors as
\[3ab(a + b) = 0.\]Thus, $a = 0,$ $b = 0,$ or $a + b = 0.$  Then $x - 3 = 0,$ $x - 7 = 0,$ or $2x - 10 = 0.$  This gives us the roots $\boxed{3, 5, 7}.$


Explanation:
Okay, so I need to figure out the side length of a regular octagon that has the same perimeter as a regular hexagon with side length 16 cm. Let me start by recalling what I know about regular polygons.

First, a regular hexagon has all sides equal and all internal angles equal. Similarly, a regular octagon also has all sides equal and all internal angles equal. The perimeter of a regular polygon is just the number of sides multiplied by the length of one side.

So, let me write down what I know:

- The hexagon has side length 16 cm.
- The perimeter of the hexagon is equal to the perimeter of the octagon.
- I need to find the side length of the octagon.

Let me start by calculating the perimeter of the hexagon. A hexagon has 6 sides. So, perimeter of hexagon (P_hex) would be 6 times the side length.

So, P_hex = 6 * 16 cm. Let me compute that:

6 * 16 = 96 cm.

So, the perimeter of the hexagon is 96 cm.

Now, the octagon has the same perimeter. A regular octagon has 8 sides. Let me denote the side length of the octagon as 's'. Therefore, the perimeter of the octagon (P_oct) is 8 * s.

Since the perimeters are equal, I can set them equal to each other:

8 * s = 96 cm.

Then, solving for 's' would be straightforward.

Divide both sides by 8:

s = 96 / 8.

Calculating that:

96 divided by 8. Well, 8*12 is 96, so s = 12 cm.

Wait, that seems straightforward. But let me check if I didn't miss anything. The problem mentions an Asymptote figure, which I can't see, but maybe there's something there? The Asymptote code draws a hexagon? Let me parse the Asymptote code to see if there's something specific.

Looking at the Asymptote code:

size(80);
pair A = dir(120);
pair B=dir(60);
pair M=(A+B)/2;
draw(dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle);
label("16 cm", M, N);

Hmm. The Asymptote code seems to be drawing a hexagon? Let me try to visualize it. The code starts at dir(360) which is the same as dir(0), so that's (1,0). Then it connects to B, which is dir(60). Then to A, which is dir(120). Then to dir(180), which is (-1,0). Then to dir(240), which is (-cos(60), -sin(60)), and then to dir(300), which is (cos(60), -sin(60)), and then back to the starting point? Wait, the draw command is: dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. So, connecting dir(360) to B (dir(60)), then to A (dir(120)), then to dir(180), then to dir(240), then to dir(300), then back to the starting point? Wait, dir(360) is the same as dir(0), so the figure is a hexagon? Let me count the sides:

Starting at dir(360) (which is (1,0)), then to dir(60), then to dir(120), then to dir(180), then to dir(240), then to dir(300), then back to dir(360). So, that's six sides. So it's a regular hexagon? Because the points are every 60 degrees? Let me check: dir(0), dir(60), dir(120), dir(180), dir(240), dir(300). Yes, those are six points on the unit circle separated by 60 degrees. So, a regular hexagon. But the label "16 cm" is placed at point M, which is the midpoint between A (dir(120)) and B (dir(60)). So, that segment AB is labeled as 16 cm.

Wait, so in the Asymptote figure, the side length of the hexagon is 16 cm? Because AB is a side of the hexagon. Let me confirm that. In the Asymptote code, the hexagon is drawn with vertices at dir(360), B, A, dir(180), dir(240), dir(300). So, the sides are between dir(360) and B, then B and A, then A and dir(180), etc. So, the segment between B and A is one side of the hexagon. Therefore, the length of AB is the side length of the hexagon, which is labeled as 16 cm. Therefore, the hexagon in the figure has side length 16 cm, which matches the problem statement.

So, there is no trick here. The problem is straightforward: perimeter of hexagon equals perimeter of octagon, so compute the side length of octagon. So, as I did before, perimeter of hexagon is 6*16=96 cm. Perimeter of octagon is 8*s, so s=96/8=12 cm.

Wait, so why was the Asymptote code given? Maybe to confirm that the hexagon is regular? Because sometimes problems might have irregular polygons, but here it's specified as a regular hexagon. So maybe just to show the figure?

Alternatively, perhaps the Asymptote code is drawing something else? Let me check again. The code is:

draw(dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle);

So starting from dir(360) which is (1,0), then to B which is dir(60), which is (cos(60°), sin(60°)) which is (0.5, sqrt(3)/2). Then to A which is dir(120), which is (-0.5, sqrt(3)/2). Then to dir(180) which is (-1,0). Then to dir(240) which is (-0.5, -sqrt(3)/2), then to dir(300) which is (0.5, -sqrt(3)/2), then back to dir(360).

So connecting these points, the figure is a regular hexagon. The label is on the segment AB, which is between dir(60) and dir(120), so that segment is a side of the hexagon. Therefore, the length of AB is 16 cm.

But in reality, in the unit circle, the distance between dir(60) and dir(120) would be the length of the side of the hexagon. Wait, but in the unit circle, the distance between two adjacent vertices is 2*sin(π/6) = 2*(1/2)=1. Wait, but if the radius is 1, then the side length is 1? Wait, no. In a regular hexagon inscribed in a circle of radius r, the side length is equal to r. Because each side subtends a 60-degree angle at the center, so the triangle formed by two radii and a side is equilateral. Therefore, the side length is equal to the radius.

But in the Asymptote code, the hexagon is drawn with vertices on the unit circle (since dir(theta) gives points on the unit circle). However, in the problem, the side length is given as 16 cm. Therefore, the Asymptote figure is just a unit circle drawing, but in reality, the side length is scaled up to 16 cm. Therefore, the figure is just for illustration, and the actual side length is 16 cm as given.

Therefore, my initial calculation is correct. The perimeter of the hexagon is 6*16=96 cm. The octagon has perimeter 8*s, so s=96/8=12 cm.

But let me check if there is any possibility that the Asymptote figure might be trying to trick me? For example, maybe the hexagon is not regular? But the problem statement says it's a regular hexagon. The Asymptote code draws a regular hexagon. So, no, that's not the case.

Alternatively, maybe the problem is in three dimensions? No, it's a regular octagon and a regular hexagon, both are 2D shapes.

Wait, another thought: sometimes, problems might refer to the perimeter in terms of the circumference or something else, but no, perimeter is straightforward.

Alternatively, could there be a mistake in the number of sides? The problem says a regular octagon, which has 8 sides. Yes.

Wait, let me just re-verify the perimeters.

For the hexagon: 6 sides, each 16 cm. So 6*16 = 96 cm.

For the octagon: 8 sides, each s cm. So 8*s.

Set equal: 8s = 96 => s = 12.

Yes. That seems correct.

Is there a possibility that the Asymptote code is drawing a different shape? Let me check again.

The Asymptote code draws dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle.

So starting from dir(360) (which is (1,0)), then to B (dir(60)), then to A (dir(120)), then to dir(180) (-1,0), then to dir(240), then to dir(300), then back to dir(360).

So the figure is a hexagon with vertices at (1,0), dir(60), dir(120), (-1,0), dir(240), dir(300).

But connecting those points, is that a regular hexagon? Let me check the distances between consecutive points.

First, between dir(360) (which is (1,0)) and B (dir(60)): distance between (1,0) and (cos(60°), sin(60°)) is sqrt[(1 - cos60)^2 + (0 - sin60)^2].

Calculating cos60° = 0.5, sin60° = sqrt(3)/2 ≈ 0.866.

So the distance squared is (1 - 0.5)^2 + ( - sqrt(3)/2)^2 = (0.5)^2 + (sqrt(3)/2)^2 = 0.25 + 0.75 = 1. Therefore, the distance is sqrt(1) = 1.

Similarly, between dir(60) and dir(120): distance between (cos60, sin60) and (cos120, sin120). cos120 = -0.5, sin120 = sqrt(3)/2.

So, difference in x: -0.5 - 0.5 = -1, difference in y: sqrt(3)/2 - sqrt(3)/2 = 0. Therefore, distance is sqrt[(-1)^2 + 0^2] = 1.

Similarly, between dir(120) and dir(180): dir(180) is (-1, 0). The distance between (-0.5, sqrt(3)/2) and (-1, 0): difference in x: -1 - (-0.5) = -0.5, difference in y: 0 - sqrt(3)/2 = -sqrt(3)/2. Squared distance: (-0.5)^2 + (-sqrt(3)/2)^2 = 0.25 + 0.75 = 1. Therefore, distance is 1.

Similarly, all sides would be length 1 in the unit circle. Therefore, in the Asymptote figure, the hexagon is regular with side length 1 (in the unit circle). But in the problem, the side length is 16 cm. Therefore, the figure is just a unit circle drawing, but scaled up so that the side length is 16 cm. Therefore, there is no issue here.

Therefore, my calculation before is correct. The side length of the octagon is 12 cm.

Wait, but to make sure, let me think again. Is there any chance that the perimeter of the octagon is different? For example, if it's a regular octagon, but in some other context? No, perimeter is always number of sides times side length.

Alternatively, maybe the problem is referring to something else? For example, in some contexts, the term "perimeter" might be confused with "circumference", but in polygons, perimeter is the total length around the polygon.

Alternatively, maybe there is a misunderstanding in the number of sides? For example, maybe the hexagon is not regular? But the problem says it's a regular hexagon.

Alternatively, maybe the Asymptote code is drawing a different figure? Let me check the Asymptote code once again. The code draws dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle.

Wait, dir(360) is (1,0). Then B is dir(60). Then A is dir(120). Then dir(180) is (-1,0). Then dir(240), then dir(300). So connecting these points. Let me check if all sides are equal. As we saw above, yes, in the unit circle, they are all length 1. So that's a regular hexagon.

Therefore, I think my answer is correct. The side length of the octagon is 12 cm.

But just to be thorough, let me do the calculation again step by step:

Perimeter of hexagon = number of sides * length of each side = 6 * 16 cm = 96 cm.

Perimeter of octagon = 8 * s, where s is the side length.

Set equal: 8s = 96 => s = 96 / 8 = 12 cm.

Yes, that's correct.

Alternatively, if I think about the relationship between the perimeters. Since the hexagon has 6 sides and the octagon has 8 sides, the octagon has 8/6 times as many sides, so to have the same perimeter, each side must be 6/8 times the length of the hexagon's side. Wait, that is:

If P_hex = P_oct,

then 6 * 16 = 8 * s => s = (6 / 8) * 16 = (3 / 4) * 16 = 12 cm.

Same result.

Alternatively, thinking in terms of ratios: the ratio of their perimeters is 1, so the ratio of their side lengths is equal to the ratio of the number of sides? No, wait, since perimeter is proportional to the side length, so if P1 = n1 * s1 and P2 = n2 * s2, then if P1 = P2, then s2 = (n1 / n2) * s1.

So here, n1 = 6 (hexagon), n2 = 8 (octagon), so s2 = (6 / 8)*s1 = (3/4)*16=12.

Same answer.

Therefore, I can be confident that the answer is 12 cm.

I think that's all. I don't see any mistakes in my reasoning. The Asymptote code just confirms that the hexagon is regular with side length 16 cm, so there's no issue there.

**Final Answer**
The length of each side of the octagon is \boxed{12} centimeters.

To solve the problem, we are given that a **regular octagon** has the **same perimeter** as a **regular hexagon** with **side length 16 cm**. Our goal is to find the **side length of the octagon**.

---

### Step 1: Understand the Shape and Its Perimeter

A **regular polygon** has all sides of equal length, and its perimeter is given by:

$$
\text{Perimeter} = \text{Number of sides} \times \text{Side length}
$$

- A **regular hexagon** has 6 sides.
- A **regular octagon** has 8 sides.

---

### Step 2: Calculate the Perimeter of the Hexagon

The side length of the hexagon is given as **16 cm**. Therefore, the perimeter of the hexagon is:

$$
P_{\text{hexagon}} = 6 \times 16 = 96 \text{ cm}
$$

---

### Step 3: Set the Perimeter Equal to the Octagon

Let $ s $ be the side length of the regular octagon. Its perimeter is:

$$
P_{\text{octagon}} = 8 \times s
$$

Since the perimeters are equal:

$$
8s = 96
$$

---

### Step 4: Solve for $ s $

$$
s = \frac{96}{8} = 12
$$

---

### Step 5: Interpret the Asymptote Code (Optional but Instructive)

The Asymptote code is provided to illustrate the shape of the hexagon. It draws a regular hexagon with vertices on the unit circle, and a segment labeled "16 cm" between two adjacent points — this confirms that the diagram visualizes the **side length** of the hexagon as **16 cm**, which aligns with the problem statement. The Asymptote drawing reinforces that we are dealing with a **regular polygon**, and there's no ambiguity in the side length.

---

### Final Answer

The length of each side of the octagon is:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the side length of a regular octagon that has the same perimeter as a regular hexagon with side length 16 cm. Let me start by recalling what I know about regular polygons.

First, a regular hexagon has all sides equal and all internal angles equal. Similarly, a regular octagon also has all sides equal and all internal angles equal. The perimeter of a regular polygon is just the number of sides multiplied by the length of one side.

So, let me write down what I know:

- The hexagon has side length 16 cm.
- The perimeter of the hexagon is equal to the perimeter of the octagon.
- I need to find the side length of the octagon.

Let me start by calculating the perimeter of the hexagon. A hexagon has 6 sides. So, perimeter of hexagon (P_hex) would be 6 times the side length.

So, P_hex = 6 * 16 cm. Let me compute that:

6 * 16 = 96 cm.

So, the perimeter of the hexagon is 96 cm.

Now, the octagon has the same perimeter. A regular octagon has 8 sides. Let me denote the side length of the octagon as 's'. Therefore, the perimeter of the octagon (P_oct) is 8 * s.

Since the perimeters are equal, I can set them equal to each other:

8 * s = 96 cm.

Then, solving for 's' would be straightforward.

Divide both sides by 8:

s = 96 / 8.

Calculating that:

96 divided by 8. Well, 8*12 is 96, so s = 12 cm.

Wait, that seems straightforward. But let me check if I didn't miss anything. The problem mentions an Asymptote figure, which I can't see, but maybe there's something there? The Asymptote code draws a hexagon? Let me parse the Asymptote code to see if there's something specific.

Looking at the Asymptote code:

size(80);
pair A = dir(120);
pair B=dir(60);
pair M=(A+B)/2;
draw(dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle);
label("16 cm", M, N);

Hmm. The Asymptote code seems to be drawing a hexagon? Let me try to visualize it. The code starts at dir(360) which is the same as dir(0), so that's (1,0). Then it connects to B, which is dir(60). Then to A, which is dir(120). Then to dir(180), which is (-1,0). Then to dir(240), which is (-cos(60), -sin(60)), and then to dir(300), which is (cos(60), -sin(60)), and then back to the starting point? Wait, the draw command is: dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. So, connecting dir(360) to B (dir(60)), then to A (dir(120)), then to dir(180), then to dir(240), then to dir(300), then back to the starting point? Wait, dir(360) is the same as dir(0), so the figure is a hexagon? Let me count the sides:

Starting at dir(360) (which is (1,0)), then to dir(60), then to dir(120), then to dir(180), then to dir(240), then to dir(300), then back to dir(360). So, that's six sides. So it's a regular hexagon? Because the points are every 60 degrees? Let me check: dir(0), dir(60), dir(120), dir(180), dir(240), dir(300). Yes, those are six points on the unit circle separated by 60 degrees. So, a regular hexagon. But the label "16 cm" is placed at point M, which is the midpoint between A (dir(120)) and B (dir(60)). So, that segment AB is labeled as 16 cm.

Wait, so in the Asymptote figure, the side length of the hexagon is 16 cm? Because AB is a side of the hexagon. Let me confirm that. In the Asymptote code, the hexagon is drawn with vertices at dir(360), B, A, dir(180), dir(240), dir(300). So, the sides are between dir(360) and B, then B and A, then A and dir(180), etc. So, the segment between B and A is one side of the hexagon. Therefore, the length of AB is the side length of the hexagon, which is labeled as 16 cm. Therefore, the hexagon in the figure has side length 16 cm, which matches the problem statement.

So, there is no trick here. The problem is straightforward: perimeter of hexagon equals perimeter of octagon, so compute the side length of octagon. So, as I did before, perimeter of hexagon is 6*16=96 cm. Perimeter of octagon is 8*s, so s=96/8=12 cm.

Wait, so why was the Asymptote code given? Maybe to confirm that the hexagon is regular? Because sometimes problems might have irregular polygons, but here it's specified as a regular hexagon. So maybe just to show the figure?

Alternatively, perhaps the Asymptote code is drawing something else? Let me check again. The code is:

draw(dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle);

So starting from dir(360) which is (1,0), then to B which is dir(60), which is (cos(60°), sin(60°)) which is (0.5, sqrt(3)/2). Then to A which is dir(120), which is (-0.5, sqrt(3)/2). Then to dir(180) which is (-1,0). Then to dir(240) which is (-0.5, -sqrt(3)/2), then to dir(300) which is (0.5, -sqrt(3)/2), then back to dir(360).

So connecting these points, the figure is a regular hexagon. The label is on the segment AB, which is between dir(60) and dir(120), so that segment is a side of the hexagon. Therefore, the length of AB is 16 cm.

But in reality, in the unit circle, the distance between dir(60) and dir(120) would be the length of the side of the hexagon. Wait, but in the unit circle, the distance between two adjacent vertices is 2*sin(π/6) = 2*(1/2)=1. Wait, but if the radius is 1, then the side length is 1? Wait, no. In a regular hexagon inscribed in a circle of radius r, the side length is equal to r. Because each side subtends a 60-degree angle at the center, so the triangle formed by two radii and a side is equilateral. Therefore, the side length is equal to the radius.

But in the Asymptote code, the hexagon is drawn with vertices on the unit circle (since dir(theta) gives points on the unit circle). However, in the problem, the side length is given as 16 cm. Therefore, the Asymptote figure is just a unit circle drawing, but in reality, the side length is scaled up to 16 cm. Therefore, the figure is just for illustration, and the actual side length is 16 cm as given.

Therefore, my initial calculation is correct. The perimeter of the hexagon is 6*16=96 cm. The octagon has perimeter 8*s, so s=96/8=12 cm.

But let me check if there is any possibility that the Asymptote figure might be trying to trick me? For example, maybe the hexagon is not regular? But the problem statement says it's a regular hexagon. The Asymptote code draws a regular hexagon. So, no, that's not the case.

Alternatively, maybe the problem is in three dimensions? No, it's a regular octagon and a regular hexagon, both are 2D shapes.

Wait, another thought: sometimes, problems might refer to the perimeter in terms of the circumference or something else, but no, perimeter is straightforward.

Alternatively, could there be a mistake in the number of sides? The problem says a regular octagon, which has 8 sides. Yes.

Wait, let me just re-verify the perimeters.

For the hexagon: 6 sides, each 16 cm. So 6*16 = 96 cm.

For the octagon: 8 sides, each s cm. So 8*s.

Set equal: 8s = 96 => s = 12.

Yes. That seems correct.

Is there a possibility that the Asymptote code is drawing a different shape? Let me check again.

The Asymptote code draws dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle.

So starting from dir(360) (which is (1,0)), then to B (dir(60)), then to A (dir(120)), then to dir(180) (-1,0), then to dir(240), then to dir(300), then back to dir(360).

So the figure is a hexagon with vertices at (1,0), dir(60), dir(120), (-1,0), dir(240), dir(300).

But connecting those points, is that a regular hexagon? Let me check the distances between consecutive points.

First, between dir(360) (which is (1,0)) and B (dir(60)): distance between (1,0) and (cos(60°), sin(60°)) is sqrt[(1 - cos60)^2 + (0 - sin60)^2].

Calculating cos60° = 0.5, sin60° = sqrt(3)/2 ≈ 0.866.

So the distance squared is (1 - 0.5)^2 + ( - sqrt(3)/2)^2 = (0.5)^2 + (sqrt(3)/2)^2 = 0.25 + 0.75 = 1. Therefore, the distance is sqrt(1) = 1.

Similarly, between dir(60) and dir(120): distance between (cos60, sin60) and (cos120, sin120). cos120 = -0.5, sin120 = sqrt(3)/2.

So, difference in x: -0.5 - 0.5 = -1, difference in y: sqrt(3)/2 - sqrt(3)/2 = 0. Therefore, distance is sqrt[(-1)^2 + 0^2] = 1.

Similarly, between dir(120) and dir(180): dir(180) is (-1, 0). The distance between (-0.5, sqrt(3)/2) and (-1, 0): difference in x: -1 - (-0.5) = -0.5, difference in y: 0 - sqrt(3)/2 = -sqrt(3)/2. Squared distance: (-0.5)^2 + (-sqrt(3)/2)^2 = 0.25 + 0.75 = 1. Therefore, distance is 1.

Similarly, all sides would be length 1 in the unit circle. Therefore, in the Asymptote figure, the hexagon is regular with side length 1 (in the unit circle). But in the problem, the side length is 16 cm. Therefore, the figure is just a unit circle drawing, but scaled up so that the side length is 16 cm. Therefore, there is no issue here.

Therefore, my calculation before is correct. The side length of the octagon is 12 cm.

Wait, but to make sure, let me think again. Is there any chance that the perimeter of the octagon is different? For example, if it's a regular octagon, but in some other context? No, perimeter is always number of sides times side length.

Alternatively, maybe the problem is referring to something else? For example, in some contexts, the term "perimeter" might be confused with "circumference", but in polygons, perimeter is the total length around the polygon.

Alternatively, maybe there is a misunderstanding in the number of sides? For example, maybe the hexagon is not regular? But the problem says it's a regular hexagon.

Alternatively, maybe the Asymptote code is drawing a different figure? Let me check the Asymptote code once again. The code draws dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle.

Wait, dir(360) is (1,0). Then B is dir(60). Then A is dir(120). Then dir(180) is (-1,0). Then dir(240), then dir(300). So connecting these points. Let me check if all sides are equal. As we saw above, yes, in the unit circle, they are all length 1. So that's a regular hexagon.

Therefore, I think my answer is correct. The side length of the octagon is 12 cm.

But just to be thorough, let me do the calculation again step by step:

Perimeter of hexagon = number of sides * length of each side = 6 * 16 cm = 96 cm.

Perimeter of octagon = 8 * s, where s is the side length.

Set equal: 8s = 96 => s = 96 / 8 = 12 cm.

Yes, that's correct.

Alternatively, if I think about the relationship between the perimeters. Since the hexagon has 6 sides and the octagon has 8 sides, the octagon has 8/6 times as many sides, so to have the same perimeter, each side must be 6/8 times the length of the hexagon's side. Wait, that is:

If P_hex = P_oct,

then 6 * 16 = 8 * s => s = (6 / 8) * 16 = (3 / 4) * 16 = 12 cm.

Same result.

Alternatively, thinking in terms of ratios: the ratio of their perimeters is 1, so the ratio of their side lengths is equal to the ratio of the number of sides? No, wait, since perimeter is proportional to the side length, so if P1 = n1 * s1 and P2 = n2 * s2, then if P1 = P2, then s2 = (n1 / n2) * s1.

So here, n1 = 6 (hexagon), n2 = 8 (octagon), so s2 = (6 / 8)*s1 = (3/4)*16=12.

Same answer.

Therefore, I can be confident that the answer is 12 cm.

I think that's all. I don't see any mistakes in my reasoning. The Asymptote code just confirms that the hexagon is regular with side length 16 cm, so there's no issue there.

**Final Answer**
The length of each side of the octagon is \boxed{12} centimeters.

To solve the problem, we are given that a **regular octagon** has the **same perimeter** as a **regular hexagon** with **side length 16 cm**. Our goal is to find the **side length of the octagon**.

---

### Step 1: Understand the Shape and Its Perimeter

A **regular polygon** has all sides of equal length, and its perimeter is given by:

$$
\text{Perimeter} = \text{Number of sides} \times \text{Side length}
$$

- A **regular hexagon** has 6 sides.
- A **regular octagon** has 8 sides.

---

### Step 2: Calculate the Perimeter of the Hexagon

The side length of the hexagon is given as **16 cm**. Therefore, the perimeter of the hexagon is:

$$
P_{\text{hexagon}} = 6 \times 16 = 96 \text{ cm}
$$

---

### Step 3: Set the Perimeter Equal to the Octagon

Let $ s $ be the side length of the regular octagon. Its perimeter is:

$$
P_{\text{octagon}} = 8 \times s
$$

Since the perimeters are equal:

$$
8s = 96
$$

---

### Step 4: Solve for $ s $

$$
s = \frac{96}{8} = 12
$$

---

### Step 5: Interpret the Asymptote Code (Optional but Instructive)

The Asymptote code is provided to illustrate the shape of the hexagon. It draws a regular hexagon with vertices on the unit circle, and a segment labeled "16 cm" between two adjacent points — this confirms that the diagram visualizes the **side length** of the hexagon as **16 cm**, which aligns with the problem statement. The Asymptote drawing reinforces that we are dealing with a **regular polygon**, and there's no ambiguity in the side length.

---

### Final Answer

The length of each side of the octagon is:

$$
\boxed{12}
$$

Solution: The hexagon has a side length of 16 centimeters, so its perimeter is $16\times 6 = 96$ centimeters. Since the octagon and the hexagon have the same perimeter, it follows that each side of the octagon has a length of $96/8 = \boxed{12}$ centimeters.


Explanation:
Okay, so I need to solve this inequality: -4 < 2(x - 1) < 8. And then express the solution in the form a < x < b, and find a + b. Alright, let me think about how to approach this.

First, inequalities can sometimes be tricky, especially when there are multiple parts. But I remember that when dealing with compound inequalities like this, I can handle each part step by step. Let me recall the properties of inequalities. One important thing is that if I divide or multiply both sides of an inequality by a positive number, the direction of the inequality sign doesn't change. But if I multiply or divide by a negative number, the direction reverses.

Looking at the given inequality: -4 < 2(x - 1) < 8. So, there's a middle term here, 2(x - 1), and it's sandwiched between -4 and 8. My goal is to solve for x.

First, I think I should isolate the term with x. That term is 2(x - 1). So, maybe I can start by dividing all parts of the inequality by 2. Let me try that.

Dividing each part by 2:

Left side: -4 divided by 2 is -2.

Middle term: 2(x - 1) divided by 2 is just (x - 1).

Right side: 8 divided by 2 is 4.

So, after dividing by 2, the inequality becomes: -2 < x - 1 < 4.

Okay, that seems simpler. Now, I need to solve for x. The next step would be to get rid of the -1 that's being subtracted from x. To do that, I can add 1 to all three parts of the inequality. Let me do that.

Adding 1 to each part:

Left side: -2 + 1 is -1.

Middle term: x - 1 + 1 is x.

Right side: 4 + 1 is 5.

So, the inequality simplifies to: -1 < x < 5.

Therefore, the solution is all real numbers x such that x is greater than -1 and less than 5. So, in the form a < x < b, a is -1 and b is 5. Therefore, a + b would be -1 + 5, which is 4.

Wait, let me check my steps again to make sure I didn't make a mistake. Let me go through the original inequality again.

Original inequality: -4 < 2(x - 1) < 8.

First step: divide all parts by 2, getting -2 < x - 1 < 4. That seems correct because dividing each part by 2 maintains the inequality.

Then, adding 1 to all parts: -2 + 1 is indeed -1, and 4 + 1 is 5. So, -1 < x < 5.

Let me verify by plugging in some numbers. Let's take x = 0, which should be in the solution. Plugging into the original inequality: 2(0 - 1) = 2(-1) = -2. So, -4 < -2 < 8. Which is true because -4 is less than -2 and -2 is less than 8.

How about x = -1? The original inequality would be 2(-1 - 1) = 2(-2) = -4. So, -4 < -4 < 8? Wait, the original inequality is -4 < ... So, -4 is not less than -4, it's equal. So, x = -1 is not included. Similarly, x = 5 would give 2(5 - 1) = 2(4) = 8, so again, the original inequality is ... < 8, so 8 is not less than 8. So, x = 5 is not included. Therefore, the solution is indeed -1 < x < 5. So, a = -1, b = 5. Therefore, a + b is -1 + 5 = 4.

Wait, but just to make sure there's no mistake in the algebra. Let me try solving it another way. Maybe expanding the original inequality first. Let's see:

Original inequality: -4 < 2(x - 1) < 8.

First, distribute the 2: -4 < 2x - 2 < 8.

Then, add 2 to all parts: -4 + 2 < 2x - 2 + 2 < 8 + 2. Which is -2 < 2x < 10.

Then divide all parts by 2: -1 < x < 5. Which is the same result as before. So that confirms the answer is correct.

Therefore, the value of a + b is -1 + 5 = 4.

But just to make sure, let me check with another value. Let's take x = 2. Then, 2(2 - 1) = 2(1) = 2. So, -4 < 2 < 8. That's true. If I take x = -2, which is less than -1. Then 2(-2 - 1) = 2(-3) = -6. Then the inequality would be -4 < -6 < 8? That's not true because -4 is greater than -6, so -4 < -6 is false. Therefore, x = -2 is not in the solution set. Similarly, x = 6 would give 2(6 -1) = 10, which is not less than 8, so x = 6 is not in the set. So the solution is indeed between -1 and 5.

Therefore, the answer is 4.

I think that's solid. The key steps were dividing by 2 first, then adding 1, and checking the endpoints. Also, confirming by expanding the original inequality gives the same result. So I feel confident that a + b is 4.

**Final Answer**
The value of $a + b$ is \boxed{4}.

To solve the inequality:

$$
-4 < 2(x - 1) < 8
$$

we aim to express the solution in the form $ a < x < b $, and then compute the value of $ a + b $.

---

### **Step 1: Simplify the Compound Inequality**

We begin by isolating the term involving $ x $:

$$
-4 < 2(x - 1) < 8
$$

Now, divide all parts of the inequality by 2 to eliminate the coefficient of the expression:

$$
\frac{-4}{2} < \frac{2(x - 1)}{2} < \frac{8}{2}
$$

$$
-2 < x - 1 < 4
$$

---

### **Step 2: Solve for $ x $**

We now solve for $ x $ by adding 1 to all parts of the inequality:

$$
-2 + 1 < x - 1 + 1 < 4 + 1
$$

$$
-1 < x < 5
$$

---

### **Step 3: Identify $ a $ and $ b $ and Compute $ a + b $**

The solution is expressed in the form $ a < x < b $, where:
- $ a = -1 $
- $ b = 5 $

Therefore, the sum:

$$
a + b = -1 + 5 = \boxed{4}
$$

---

### ✅ **Final Answer**

$$
\boxed{4}
$$

[Assistant]
Explanation:
Okay, so I need to solve this inequality: -4 < 2(x - 1) < 8. And then express the solution in the form a < x < b, and find a + b. Alright, let me think about how to approach this.

First, inequalities can sometimes be tricky, especially when there are multiple parts. But I remember that when dealing with compound inequalities like this, I can handle each part step by step. Let me recall the properties of inequalities. One important thing is that if I divide or multiply both sides of an inequality by a positive number, the direction of the inequality sign doesn't change. But if I multiply or divide by a negative number, the direction reverses.

Looking at the given inequality: -4 < 2(x - 1) < 8. So, there's a middle term here, 2(x - 1), and it's sandwiched between -4 and 8. My goal is to solve for x.

First, I think I should isolate the term with x. That term is 2(x - 1). So, maybe I can start by dividing all parts of the inequality by 2. Let me try that.

Dividing each part by 2:

Left side: -4 divided by 2 is -2.

Middle term: 2(x - 1) divided by 2 is just (x - 1).

Right side: 8 divided by 2 is 4.

So, after dividing by 2, the inequality becomes: -2 < x - 1 < 4.

Okay, that seems simpler. Now, I need to solve for x. The next step would be to get rid of the -1 that's being subtracted from x. To do that, I can add 1 to all three parts of the inequality. Let me do that.

Adding 1 to each part:

Left side: -2 + 1 is -1.

Middle term: x - 1 + 1 is x.

Right side: 4 + 1 is 5.

So, the inequality simplifies to: -1 < x < 5.

Therefore, the solution is all real numbers x such that x is greater than -1 and less than 5. So, in the form a < x < b, a is -1 and b is 5. Therefore, a + b would be -1 + 5, which is 4.

Wait, let me check my steps again to make sure I didn't make a mistake. Let me go through the original inequality again.

Original inequality: -4 < 2(x - 1) < 8.

First step: divide all parts by 2, getting -2 < x - 1 < 4. That seems correct because dividing each part by 2 maintains the inequality.

Then, adding 1 to all parts: -2 + 1 is indeed -1, and 4 + 1 is 5. So, -1 < x < 5.

Let me verify by plugging in some numbers. Let's take x = 0, which should be in the solution. Plugging into the original inequality: 2(0 - 1) = 2(-1) = -2. So, -4 < -2 < 8. Which is true because -4 is less than -2 and -2 is less than 8.

How about x = -1? The original inequality would be 2(-1 - 1) = 2(-2) = -4. So, -4 < -4 < 8? Wait, the original inequality is -4 < ... So, -4 is not less than -4, it's equal. So, x = -1 is not included. Similarly, x = 5 would give 2(5 - 1) = 2(4) = 8, so again, the original inequality is ... < 8, so 8 is not less than 8. So, x = 5 is not included. Therefore, the solution is indeed -1 < x < 5. So, a = -1, b = 5. Therefore, a + b is -1 + 5 = 4.

Wait, but just to make sure there's no mistake in the algebra. Let me try solving it another way. Maybe expanding the original inequality first. Let's see:

Original inequality: -4 < 2(x - 1) < 8.

First, distribute the 2: -4 < 2x - 2 < 8.

Then, add 2 to all parts: -4 + 2 < 2x - 2 + 2 < 8 + 2. Which is -2 < 2x < 10.

Then divide all parts by 2: -1 < x < 5. Which is the same result as before. So that confirms the answer is correct.

Therefore, the value of a + b is -1 + 5 = 4.

But just to make sure, let me check with another value. Let's take x = 2. Then, 2(2 - 1) = 2(1) = 2. So, -4 < 2 < 8. That's true. If I take x = -2, which is less than -1. Then 2(-2 - 1) = 2(-3) = -6. Then the inequality would be -4 < -6 < 8? That's not true because -4 is greater than -6, so -4 < -6 is false. Therefore, x = -2 is not in the solution set. Similarly, x = 6 would give 2(6 -1) = 10, which is not less than 8, so x = 6 is not in the set. So the solution is indeed between -1 and 5.

Therefore, the answer is 4.

I think that's solid. The key steps were dividing by 2 first, then adding 1, and checking the endpoints. Also, confirming by expanding the original inequality gives the same result. So I feel confident that a + b is 4.

**Final Answer**
The value of $a + b$ is \boxed{4}.

To solve the inequality:

$$
-4 < 2(x - 1) < 8
$$

we aim to express the solution in the form $ a < x < b $, and then compute the value of $ a + b $.

---

### **Step 1: Simplify the Compound Inequality**

We begin by isolating the term involving $ x $:

$$
-4 < 2(x - 1) < 8
$$

Now, divide all parts of the inequality by 2 to eliminate the coefficient of the expression:

$$
\frac{-4}{2} < \frac{2(x - 1)}{2} < \frac{8}{2}
$$

$$
-2 < x - 1 < 4
$$

---

### **Step 2: Solve for $ x $**

We now solve for $ x $ by adding 1 to all parts of the inequality:

$$
-2 + 1 < x - 1 + 1 < 4 + 1
$$

$$
-1 < x < 5
$$

---

### **Step 3: Identify $ a $ and $ b $ and Compute $ a + b $**

The solution is expressed in the form $ a < x < b $, where:
- $ a = -1 $
- $ b = 5 $

Therefore, the sum:

$$
a + b = -1 + 5 = \boxed{4}
$$

---

### ✅ **Final Answer**

$$
\boxed{4}
$$

Solution: Since everything in sight is even, we should begin by dividing by 2.  That gives \[-2<x-1<4.\] To isolate $x$, we add 1, so \[-1<x<5.\] Since $a=-1$ and $b=5$, we get $a+b=-1+5=\boxed{4}$.