Explanation:
Okay, so I need to figure out how many different battalions can be sent to the Battle of Thermopylae based on the number of upper and lower class soldiers available. Let me start by understanding the problem.

First, the Greek army has two types of soldiers: upper class and lower class. In a certain part of Athens, there are 5 upper class soldiers and 10 lower class soldiers. The battle requires a force of 4 upper class soldiers and 8 lower class soldiers. The question is asking how many different battalions can be sent.

Hmm, so I think this is a combinatorics problem where I need to calculate the number of ways to choose the required number of soldiers from each class. Since the battle requires specific numbers of each type, I need to find the number of combinations for upper class soldiers and lower class soldiers separately and then multiply them together because the choices are independent.

Let me confirm: the problem says "different battalions," so each battalion is a group of soldiers that meets the required numbers. Since the battle demands exactly 4 upper class and 8 lower class soldiers, each battalion must consist of exactly those numbers. Therefore, the question is essentially asking how many ways we can choose 4 upper class soldiers out of 5 and 8 lower class soldiers out of 10. Then, the total number of different battalions would be the product of these two combinations.

So, first, let me recall the combination formula. The number of ways to choose k elements from a set of n elements is given by the combination formula C(n, k) = n! / (k! * (n - k)!).

Therefore, for the upper class soldiers: we have 5 soldiers and need to choose 4. So that would be C(5, 4). Let me compute that.

C(5, 4) = 5! / (4! * (5 - 4)!) = (5 * 4 * 3 * 2 * 1) / [(4 * 3 * 2 * 1) * (1)] = 5 / 1 = 5. So there are 5 ways to choose the upper class soldiers.

Now for the lower class soldiers: there are 10 soldiers and we need to choose 8. So that's C(10, 8). Let me calculate that.

C(10, 8) is the same as C(10, 2) because of the combination symmetry property C(n, k) = C(n, n - k). So C(10, 2) = 10! / (2! * 8!) = (10 * 9 * 8! ) / (2 * 1 * 8!) ) = (10 * 9)/2 = 90 / 2 = 45. Therefore, there are 45 ways to choose the lower class soldiers.

Therefore, the total number of different battalions is the product of these two numbers: 5 * 45. Let me compute that. 5 * 45 is 225.

Wait, but let me make sure I didn't make any mistakes here. Let me check the combination calculations again.

For upper class: 5 choose 4. Yes, that's 5. Because choosing 4 out of 5 is the same as leaving out 1, and there are 5 choices for who is left out. So that's correct.

For lower class: 10 choose 8. Which is 45. Let me verify with the formula. 10! / (8! * 2!) = (10 * 9 * 8! ) / (8! * 2 * 1) ) = (10 * 9)/2 = 45. Yes, that's correct. So 45 ways.

Therefore, multiplying them together: 5 * 45 = 225. So there are 225 different battalions that can be sent.

But wait, let me think again. Is there any possibility that the problem is asking for something else? For example, maybe the battalions are considered different if they have different numbers of soldiers? But no, the problem says the battle demands a force of 4 upper and 8 lower. So each battalion must have exactly 4 upper and 8 lower. Therefore, the number of different battalions is the number of ways to choose those soldiers.

Alternatively, could there be a different interpretation? For example, maybe the question is about forming battalions of any size that meet the requirement? But no, the battle demands a specific force, so the battalion must be exactly that size. Therefore, the answer is 5 * 45 = 225.

Alternatively, maybe the problem is in the translation? The original says "how many different battalions can be sent?" If "battalion" refers to a group that meets the required numbers, then yes, it's the product.

Alternatively, maybe there's a restriction that a battalion must consist of soldiers from both classes? But since the battle requires both upper and lower class soldiers, and we are choosing exactly 4 and 8, so each battalion will have both classes. So no problem there.

Alternatively, maybe the problem is considering that different orders of soldiers in the battalion count as different? But no, in combinatorics, unless specified, we assume that the order doesn't matter. So combinations are the right approach here.

Therefore, I think my reasoning is correct. The answer is 225.

But let me check once more. Suppose there are 5 upper class soldiers. Let's name them A, B, C, D, E. Choosing 4 of them: the possible combinations are all except each one individually. So, omitting A: B, C, D, E; omitting B: A, C, D, E; etc. So 5 combinations. Correct.

For lower class soldiers, 10 soldiers. Choosing 8 is equivalent to leaving out 2. The number of ways to leave out 2 soldiers from 10 is C(10, 2) = 45. So that's correct.

Therefore, total number of battalions is 5 * 45 = 225.

Therefore, I think the answer is 225.

**Final Answer**
The number of different battalions that can be sent is \boxed{225}.

To determine how many different battalions can be sent to the Battle of Thermopylae, we need to analyze the composition of the available soldiers and the specific requirements of the battle.

---

### **Step 1: Understand the Composition of the Army**

- **Upper class soldiers**: 5 available
- **Lower class soldiers**: 10 available

---

### **Step 2: Determine the Battle Requirements**

The battle requires:

- **4 upper class soldiers**
- **8 lower class soldiers**

Each battalion must consist of exactly these numbers of soldiers.

---

### **Step 3: Calculate the Number of Ways to Choose Soldiers**

We use the **combination formula** to determine how many ways we can choose the required number of soldiers from each class:

$$
C(n, k) = \frac{n!}{k!(n-k)!}
$$

#### **Upper Class Soldiers:**
We need to choose 4 out of 5 upper class soldiers:

$$
C(5, 4) = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = 5
$$

So, there are **5** ways to choose 4 upper class soldiers.

#### **Lower Class Soldiers:**
We need to choose 8 out of 10 lower class soldiers:

$$
C(10, 8) = C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45
$$

So, there are **45** ways to choose 8 lower class soldiers.

---

### **Step 4: Calculate the Total Number of Different Battalions**

Since the choices of upper and lower class soldiers are independent, the total number of different battalions is the product of the two combinations:

$$
\text{Total Battalions} = C(5, 4) \times C(10, 8) = 5 \times 45 = 225
$$

---

### **Final Answer**

$$
\boxed{225}
$$

[Assistant]
Explanation:
Okay, so I need to figure out how many different battalions can be sent to the Battle of Thermopylae based on the number of upper and lower class soldiers available. Let me start by understanding the problem.

First, the Greek army has two types of soldiers: upper class and lower class. In a certain part of Athens, there are 5 upper class soldiers and 10 lower class soldiers. The battle requires a force of 4 upper class soldiers and 8 lower class soldiers. The question is asking how many different battalions can be sent.

Hmm, so I think this is a combinatorics problem where I need to calculate the number of ways to choose the required number of soldiers from each class. Since the battle requires specific numbers of each type, I need to find the number of combinations for upper class soldiers and lower class soldiers separately and then multiply them together because the choices are independent.

Let me confirm: the problem says "different battalions," so each battalion is a group of soldiers that meets the required numbers. Since the battle demands exactly 4 upper class and 8 lower class soldiers, each battalion must consist of exactly those numbers. Therefore, the question is essentially asking how many ways we can choose 4 upper class soldiers out of 5 and 8 lower class soldiers out of 10. Then, the total number of different battalions would be the product of these two combinations.

So, first, let me recall the combination formula. The number of ways to choose k elements from a set of n elements is given by the combination formula C(n, k) = n! / (k! * (n - k)!).

Therefore, for the upper class soldiers: we have 5 soldiers and need to choose 4. So that would be C(5, 4). Let me compute that.

C(5, 4) = 5! / (4! * (5 - 4)!) = (5 * 4 * 3 * 2 * 1) / [(4 * 3 * 2 * 1) * (1)] = 5 / 1 = 5. So there are 5 ways to choose the upper class soldiers.

Now for the lower class soldiers: there are 10 soldiers and we need to choose 8. So that's C(10, 8). Let me calculate that.

C(10, 8) is the same as C(10, 2) because of the combination symmetry property C(n, k) = C(n, n - k). So C(10, 2) = 10! / (2! * 8!) = (10 * 9 * 8! ) / (2 * 1 * 8!) ) = (10 * 9)/2 = 90 / 2 = 45. Therefore, there are 45 ways to choose the lower class soldiers.

Therefore, the total number of different battalions is the product of these two numbers: 5 * 45. Let me compute that. 5 * 45 is 225.

Wait, but let me make sure I didn't make any mistakes here. Let me check the combination calculations again.

For upper class: 5 choose 4. Yes, that's 5. Because choosing 4 out of 5 is the same as leaving out 1, and there are 5 choices for who is left out. So that's correct.

For lower class: 10 choose 8. Which is 45. Let me verify with the formula. 10! / (8! * 2!) = (10 * 9 * 8! ) / (8! * 2 * 1) ) = (10 * 9)/2 = 45. Yes, that's correct. So 45 ways.

Therefore, multiplying them together: 5 * 45 = 225. So there are 225 different battalions that can be sent.

But wait, let me think again. Is there any possibility that the problem is asking for something else? For example, maybe the battalions are considered different if they have different numbers of soldiers? But no, the problem says the battle demands a force of 4 upper and 8 lower. So each battalion must have exactly 4 upper and 8 lower. Therefore, the number of different battalions is the number of ways to choose those soldiers.

Alternatively, could there be a different interpretation? For example, maybe the question is about forming battalions of any size that meet the requirement? But no, the battle demands a specific force, so the battalion must be exactly that size. Therefore, the answer is 5 * 45 = 225.

Alternatively, maybe the problem is in the translation? The original says "how many different battalions can be sent?" If "battalion" refers to a group that meets the required numbers, then yes, it's the product.

Alternatively, maybe there's a restriction that a battalion must consist of soldiers from both classes? But since the battle requires both upper and lower class soldiers, and we are choosing exactly 4 and 8, so each battalion will have both classes. So no problem there.

Alternatively, maybe the problem is considering that different orders of soldiers in the battalion count as different? But no, in combinatorics, unless specified, we assume that the order doesn't matter. So combinations are the right approach here.

Therefore, I think my reasoning is correct. The answer is 225.

But let me check once more. Suppose there are 5 upper class soldiers. Let's name them A, B, C, D, E. Choosing 4 of them: the possible combinations are all except each one individually. So, omitting A: B, C, D, E; omitting B: A, C, D, E; etc. So 5 combinations. Correct.

For lower class soldiers, 10 soldiers. Choosing 8 is equivalent to leaving out 2. The number of ways to leave out 2 soldiers from 10 is C(10, 2) = 45. So that's correct.

Therefore, total number of battalions is 5 * 45 = 225.

Therefore, I think the answer is 225.

**Final Answer**
The number of different battalions that can be sent is \boxed{225}.

To determine how many different battalions can be sent to the Battle of Thermopylae, we need to analyze the composition of the available soldiers and the specific requirements of the battle.

---

### **Step 1: Understand the Composition of the Army**

- **Upper class soldiers**: 5 available
- **Lower class soldiers**: 10 available

---

### **Step 2: Determine the Battle Requirements**

The battle requires:

- **4 upper class soldiers**
- **8 lower class soldiers**

Each battalion must consist of exactly these numbers of soldiers.

---

### **Step 3: Calculate the Number of Ways to Choose Soldiers**

We use the **combination formula** to determine how many ways we can choose the required number of soldiers from each class:

$$
C(n, k) = \frac{n!}{k!(n-k)!}
$$

#### **Upper Class Soldiers:**
We need to choose 4 out of 5 upper class soldiers:

$$
C(5, 4) = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = 5
$$

So, there are **5** ways to choose 4 upper class soldiers.

#### **Lower Class Soldiers:**
We need to choose 8 out of 10 lower class soldiers:

$$
C(10, 8) = C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45
$$

So, there are **45** ways to choose 8 lower class soldiers.

---

### **Step 4: Calculate the Total Number of Different Battalions**

Since the choices of upper and lower class soldiers are independent, the total number of different battalions is the product of the two combinations:

$$
\text{Total Battalions} = C(5, 4) \times C(10, 8) = 5 \times 45 = 225
$$

---

### **Final Answer**

$$
\boxed{225}
$$

Solution: There are $\binom{5}{4}$ different ways to choose 4 from 5 upper class soldiers. For each of these, there are $\binom{10}{8}$ ways to choose 8 lower class soldiers. The number of different battalions, then, is $\binom{5}{4}\cdot \binom{10}{8} = \boxed{225}$.


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.

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, so that means n can be written as 7k + 2 for some integer k. But maybe instead of working with variables, I can just replace n with 2 in the expression because when we take modulo 7, the value of n is equivalent to 2. Let me check if that's valid.

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

So substituting n = 2:

First term: n + 2 = 2 + 2 = 4

Second term: n + 4 = 2 + 4 = 6

Third term: n + 6 = 2 + 6 = 8

Now, multiply them together: 4 * 6 * 8. Let me compute that step by step.

First, 4 * 6 = 24. Then, 24 * 8. Let me calculate 24 * 8. Well, 20*8=160 and 4*8=32, so 160 + 32 = 192. So the product is 192. Now, I need to find the remainder when 192 is divided by 7.

Alternatively, since we're working modulo 7, maybe I can compute each term modulo 7 first and then multiply them. Let me try that approach as well to verify.

Original terms:

n + 2 ≡ 2 + 2 = 4 mod 7

n + 4 ≡ 2 + 4 = 6 mod 7

n + 6 ≡ 2 + 6 = 8 mod 7. But 8 mod 7 is 1, since 7*1=7 and 8-7=1. So that term is 1 mod 7.

Therefore, the product is 4 * 6 * 1 mod 7. Let me compute that.

First, 4 * 6 = 24. Then 24 mod 7. Let me divide 24 by 7. 7*3=21, so 24 - 21 = 3. Therefore, 24 mod 7 is 3. Then multiply by 1: 3 * 1 = 3 mod 7. So the remainder is 3.

Wait, but earlier when I calculated 192 divided by 7, let me check that. Let me divide 192 by 7. 7*27=189, so 192 - 189 = 3. So yes, 192 mod 7 is 3. So both methods give me the same result. Therefore, the remainder is 3.

But let me think again if there's another way to approach this problem, maybe using properties of modular arithmetic or something else. For example, since n ≡ 2 mod 7, then n is congruent to 2, so maybe we can think of the three terms (n + 2), (n + 4), (n + 6) as numbers that are 4, 6, 8 mod 7, which simplifies to 4, 6, 1 mod 7. Then multiplying them gives 4*6*1=24≡3 mod7.

Alternatively, maybe there's a pattern here. Let me think about the three consecutive even numbers? Wait, n + 2, n + 4, n + 6 are three consecutive numbers with a difference of 2. But since we are working modulo 7, maybe they are three numbers spaced by 2. However, since 7 is a prime number, maybe there's some properties here.

Alternatively, maybe I can think of the product (n + 2)(n + 4)(n + 6) as a polynomial in n. Let me expand it.

First, let me compute (n + 2)(n + 4). Let me multiply those two first.

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

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

But maybe expanding isn't helpful here because we need to compute this modulo 7. However, since n ≡ 2 mod 7, we can substitute n = 2 into this polynomial. Let me try that.

So substituting n = 2:

n^3 = 8

12n^2 = 12*(4) = 48

44n = 44*2 = 88

48 is just 48.

Adding them all up: 8 + 48 + 88 + 48. Let me compute that:

First, 8 + 48 = 56

Then 56 + 88 = 144

Then 144 + 48 = 192. Which is the same as before. Then 192 mod 7 is 3. So same result.

Alternatively, maybe I can compute each term modulo 7 first. Let me try that.

n ≡ 2 mod7, so:

n^3 ≡ 2^3 = 8 ≡ 1 mod7 (since 8 - 7 = 1)

12n^2: 12 mod7 is 5, since 7*1=7, 12-7=5. Then n^2 ≡ 2^2 = 4 mod7. So 5*4 = 20 ≡ 20 - 14 = 6 mod7.

44n: 44 mod7. Let me divide 44 by 7. 7*6=42, so 44 - 42 = 2. So 44 ≡ 2 mod7. Then n ≡ 2 mod7, so 2*2 = 4 mod7.

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

Now adding all these together: 1 (from n^3) + 6 (from 12n^2) + 4 (from 44n) + 6 (from 48) = 1 + 6 + 4 + 6 = 17. Then 17 mod7: 7*2=14, 17 -14=3. So again, 3 mod7.

So regardless of the method, I get 3. Therefore, the remainder is 3.

But let me check if there's another way to think about this. For example, since n ≡ 2 mod7, then n is 2 more than a multiple of 7. So, when we add 2, 4, 6 to n, we get numbers that are 4, 6, 8 mod7. But 8 mod7 is 1, so the three terms are 4, 6, 1. So multiplying them: 4*6*1.

Alternatively, maybe I can think of these numbers as residues modulo 7. Let me list the residues:

If n ≡ 2 mod7, then:

n + 2 ≡ 4 mod7

n + 4 ≡ 6 mod7

n + 6 ≡ 8 ≡ 1 mod7

So the product is 4 * 6 * 1. Let me compute 4*6 first. 4*6=24. 24 mod7 is 3, as before. Then 3*1=3.

Alternatively, maybe there's a pattern here. Let me think about the product of three numbers that are spaced by 2. For example, if n ≡ 2 mod7, then the three numbers are 2 + 2, 2 + 4, 2 + 6. Which is 4, 6, 8. But 8 is 1 mod7. So 4, 6, 1.

Alternatively, maybe I can think of these numbers as 4, 6, 1. Let me check if there's a relationship between them. For example, 4*6*1 = 24. But 24 mod7 is 3.

Alternatively, maybe I can think of the product (n + 2)(n + 4)(n + 6) as something related to factorials or something else. But maybe not necessary here.

Alternatively, maybe I can think of the three terms as consecutive numbers in some sense. Wait, 4, 6, 1. If we consider modulo 7, they are not consecutive numbers, but they are spaced by 2.

Alternatively, maybe I can think of the product (n + 2)(n + 4)(n + 6) as (n + 2)(n + 4)(n + 6) mod7. Since n ≡2 mod7, then substituting n=2 gives the same result.

Alternatively, maybe I can think of the problem in terms of variables. Let me suppose that n ≡2 mod7, so let me write n = 7k + 2. Then compute (n + 2)(n + 4)(n + 6):

First, n + 2 = 7k + 2 + 2 = 7k + 4

n + 4 = 7k + 2 + 4 = 7k + 6

n + 6 = 7k + 2 + 6 = 7k + 8

So the product is (7k + 4)(7k + 6)(7k + 8). Now, if I expand this, it's going to be a cubic polynomial in k. However, since we are interested in the value modulo 7, all terms with 7k will be multiples of 7, so when we take modulo 7, they will vanish. Therefore, the entire expression modulo 7 is equivalent to (0 + 4)(0 + 6)(0 + 8) mod7, which is 4*6*8 mod7. Which is the same as before.

But 8 mod7 is 1, so again 4*6*1 = 24 mod7 = 3.

Therefore, regardless of the approach, the answer is 3.

Wait, but let me check with a concrete example. Let me pick a number n that is congruent to 2 mod7, say n=2. Then compute (2 + 2)(2 + 4)(2 + 6) = 4*6*8 = 192. 192 divided by 7 is 27*7=189, remainder 3. So yes, remainder 3.

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

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

Another example: n=16 (since 16 mod7 is 2). Then compute (16 + 2)(16 + 4)(16 + 6) = 18*20*22. Let me compute 18*20=360, then 360*22. 360*20=7200, 360*2=720, total 7200 + 720 = 7920. Now divide 7920 by 7. Let me see: 7*1100=7700. 7920 - 7700 = 220. 220 divided by 7: 7*31=217, remainder 3. Again, remainder 3.

So all examples confirm that the remainder is 3. Therefore, I can be confident that the answer is 3.

But just to make sure, let me think if there's any possible mistake in my reasoning. For example, could there be a case where substituting n with 2 mod7 doesn't hold? Well, in modular arithmetic, if n ≡ a mod m, then any polynomial expression in n will be congruent to the same polynomial evaluated at a mod m. So substituting n with 2 is valid here.

Alternatively, maybe I can think of the problem in terms of the multiplicative inverses or something else, but I don't think that's necessary here.

Alternatively, maybe there's a general formula for such products. For example, if we have three consecutive even numbers or something. But since we are working modulo 7, and the numbers are spaced by 2, maybe there's a pattern.

But since all methods lead to the same answer, and examples confirm it, I think the answer is definitely 3.

Therefore, the remainder when (n + 2)(n + 4)(n + 6) is divided by 7 is 3.

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

To solve the problem, we are given that:

$$
n \equiv 2 \pmod{7}
$$

We are asked to find the remainder when the expression:

$$
(n + 2)(n + 4)(n + 6)
$$

is divided by 7.

---

### Step 1: Use the Given Congruence

Since $ n \equiv 2 \pmod{7} $, we can substitute $ n = 2 $ into the expression. This is valid because in modular arithmetic, if $ n \equiv a \pmod{m} $, then any expression involving $ n $ can be evaluated by replacing $ n $ with $ a $, modulo $ m $.

So, we compute:

$$
(n + 2)(n + 4)(n + 6) \equiv (2 + 2)(2 + 4)(2 + 6) \pmod{7}
$$

$$
\equiv (4)(6)(8) \pmod{7}
$$

---

### Step 2: Simplify Each Term Modulo 7

Now simplify each term modulo 7:

- $ 4 \mod 7 = 4 $
- $ 6 \mod 7 = 6 $
- $ 8 \mod 7 = 1 $

So the expression becomes:

$$
(4)(6)(1) \pmod{7}
$$

---

### Step 3: Multiply and Reduce Modulo 7

First, compute:

$$
4 \times 6 = 24
$$

Then:

$$
24 \mod 7 = 3
$$

So the final result is:

$$
3 \pmod{7}
$$

---

### Final Answer

$$
\boxed{3}
$$

[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.

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, so that means n can be written as 7k + 2 for some integer k. But maybe instead of working with variables, I can just replace n with 2 in the expression because when we take modulo 7, the value of n is equivalent to 2. Let me check if that's valid.

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

So substituting n = 2:

First term: n + 2 = 2 + 2 = 4

Second term: n + 4 = 2 + 4 = 6

Third term: n + 6 = 2 + 6 = 8

Now, multiply them together: 4 * 6 * 8. Let me compute that step by step.

First, 4 * 6 = 24. Then, 24 * 8. Let me calculate 24 * 8. Well, 20*8=160 and 4*8=32, so 160 + 32 = 192. So the product is 192. Now, I need to find the remainder when 192 is divided by 7.

Alternatively, since we're working modulo 7, maybe I can compute each term modulo 7 first and then multiply them. Let me try that approach as well to verify.

Original terms:

n + 2 ≡ 2 + 2 = 4 mod 7

n + 4 ≡ 2 + 4 = 6 mod 7

n + 6 ≡ 2 + 6 = 8 mod 7. But 8 mod 7 is 1, since 7*1=7 and 8-7=1. So that term is 1 mod 7.

Therefore, the product is 4 * 6 * 1 mod 7. Let me compute that.

First, 4 * 6 = 24. Then 24 mod 7. Let me divide 24 by 7. 7*3=21, so 24 - 21 = 3. Therefore, 24 mod 7 is 3. Then multiply by 1: 3 * 1 = 3 mod 7. So the remainder is 3.

Wait, but earlier when I calculated 192 divided by 7, let me check that. Let me divide 192 by 7. 7*27=189, so 192 - 189 = 3. So yes, 192 mod 7 is 3. So both methods give me the same result. Therefore, the remainder is 3.

But let me think again if there's another way to approach this problem, maybe using properties of modular arithmetic or something else. For example, since n ≡ 2 mod 7, then n is congruent to 2, so maybe we can think of the three terms (n + 2), (n + 4), (n + 6) as numbers that are 4, 6, 8 mod 7, which simplifies to 4, 6, 1 mod 7. Then multiplying them gives 4*6*1=24≡3 mod7.

Alternatively, maybe there's a pattern here. Let me think about the three consecutive even numbers? Wait, n + 2, n + 4, n + 6 are three consecutive numbers with a difference of 2. But since we are working modulo 7, maybe they are three numbers spaced by 2. However, since 7 is a prime number, maybe there's some properties here.

Alternatively, maybe I can think of the product (n + 2)(n + 4)(n + 6) as a polynomial in n. Let me expand it.

First, let me compute (n + 2)(n + 4). Let me multiply those two first.

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

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

But maybe expanding isn't helpful here because we need to compute this modulo 7. However, since n ≡ 2 mod 7, we can substitute n = 2 into this polynomial. Let me try that.

So substituting n = 2:

n^3 = 8

12n^2 = 12*(4) = 48

44n = 44*2 = 88

48 is just 48.

Adding them all up: 8 + 48 + 88 + 48. Let me compute that:

First, 8 + 48 = 56

Then 56 + 88 = 144

Then 144 + 48 = 192. Which is the same as before. Then 192 mod 7 is 3. So same result.

Alternatively, maybe I can compute each term modulo 7 first. Let me try that.

n ≡ 2 mod7, so:

n^3 ≡ 2^3 = 8 ≡ 1 mod7 (since 8 - 7 = 1)

12n^2: 12 mod7 is 5, since 7*1=7, 12-7=5. Then n^2 ≡ 2^2 = 4 mod7. So 5*4 = 20 ≡ 20 - 14 = 6 mod7.

44n: 44 mod7. Let me divide 44 by 7. 7*6=42, so 44 - 42 = 2. So 44 ≡ 2 mod7. Then n ≡ 2 mod7, so 2*2 = 4 mod7.

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

Now adding all these together: 1 (from n^3) + 6 (from 12n^2) + 4 (from 44n) + 6 (from 48) = 1 + 6 + 4 + 6 = 17. Then 17 mod7: 7*2=14, 17 -14=3. So again, 3 mod7.

So regardless of the method, I get 3. Therefore, the remainder is 3.

But let me check if there's another way to think about this. For example, since n ≡ 2 mod7, then n is 2 more than a multiple of 7. So, when we add 2, 4, 6 to n, we get numbers that are 4, 6, 8 mod7. But 8 mod7 is 1, so the three terms are 4, 6, 1. So multiplying them: 4*6*1.

Alternatively, maybe I can think of these numbers as residues modulo 7. Let me list the residues:

If n ≡ 2 mod7, then:

n + 2 ≡ 4 mod7

n + 4 ≡ 6 mod7

n + 6 ≡ 8 ≡ 1 mod7

So the product is 4 * 6 * 1. Let me compute 4*6 first. 4*6=24. 24 mod7 is 3, as before. Then 3*1=3.

Alternatively, maybe there's a pattern here. Let me think about the product of three numbers that are spaced by 2. For example, if n ≡ 2 mod7, then the three numbers are 2 + 2, 2 + 4, 2 + 6. Which is 4, 6, 8. But 8 is 1 mod7. So 4, 6, 1.

Alternatively, maybe I can think of these numbers as 4, 6, 1. Let me check if there's a relationship between them. For example, 4*6*1 = 24. But 24 mod7 is 3.

Alternatively, maybe I can think of the product (n + 2)(n + 4)(n + 6) as something related to factorials or something else. But maybe not necessary here.

Alternatively, maybe I can think of the three terms as consecutive numbers in some sense. Wait, 4, 6, 1. If we consider modulo 7, they are not consecutive numbers, but they are spaced by 2.

Alternatively, maybe I can think of the product (n + 2)(n + 4)(n + 6) as (n + 2)(n + 4)(n + 6) mod7. Since n ≡2 mod7, then substituting n=2 gives the same result.

Alternatively, maybe I can think of the problem in terms of variables. Let me suppose that n ≡2 mod7, so let me write n = 7k + 2. Then compute (n + 2)(n + 4)(n + 6):

First, n + 2 = 7k + 2 + 2 = 7k + 4

n + 4 = 7k + 2 + 4 = 7k + 6

n + 6 = 7k + 2 + 6 = 7k + 8

So the product is (7k + 4)(7k + 6)(7k + 8). Now, if I expand this, it's going to be a cubic polynomial in k. However, since we are interested in the value modulo 7, all terms with 7k will be multiples of 7, so when we take modulo 7, they will vanish. Therefore, the entire expression modulo 7 is equivalent to (0 + 4)(0 + 6)(0 + 8) mod7, which is 4*6*8 mod7. Which is the same as before.

But 8 mod7 is 1, so again 4*6*1 = 24 mod7 = 3.

Therefore, regardless of the approach, the answer is 3.

Wait, but let me check with a concrete example. Let me pick a number n that is congruent to 2 mod7, say n=2. Then compute (2 + 2)(2 + 4)(2 + 6) = 4*6*8 = 192. 192 divided by 7 is 27*7=189, remainder 3. So yes, remainder 3.

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

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

Another example: n=16 (since 16 mod7 is 2). Then compute (16 + 2)(16 + 4)(16 + 6) = 18*20*22. Let me compute 18*20=360, then 360*22. 360*20=7200, 360*2=720, total 7200 + 720 = 7920. Now divide 7920 by 7. Let me see: 7*1100=7700. 7920 - 7700 = 220. 220 divided by 7: 7*31=217, remainder 3. Again, remainder 3.

So all examples confirm that the remainder is 3. Therefore, I can be confident that the answer is 3.

But just to make sure, let me think if there's any possible mistake in my reasoning. For example, could there be a case where substituting n with 2 mod7 doesn't hold? Well, in modular arithmetic, if n ≡ a mod m, then any polynomial expression in n will be congruent to the same polynomial evaluated at a mod m. So substituting n with 2 is valid here.

Alternatively, maybe I can think of the problem in terms of the multiplicative inverses or something else, but I don't think that's necessary here.

Alternatively, maybe there's a general formula for such products. For example, if we have three consecutive even numbers or something. But since we are working modulo 7, and the numbers are spaced by 2, maybe there's a pattern.

But since all methods lead to the same answer, and examples confirm it, I think the answer is definitely 3.

Therefore, the remainder when (n + 2)(n + 4)(n + 6) is divided by 7 is 3.

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

To solve the problem, we are given that:

$$
n \equiv 2 \pmod{7}
$$

We are asked to find the remainder when the expression:

$$
(n + 2)(n + 4)(n + 6)
$$

is divided by 7.

---

### Step 1: Use the Given Congruence

Since $ n \equiv 2 \pmod{7} $, we can substitute $ n = 2 $ into the expression. This is valid because in modular arithmetic, if $ n \equiv a \pmod{m} $, then any expression involving $ n $ can be evaluated by replacing $ n $ with $ a $, modulo $ m $.

So, we compute:

$$
(n + 2)(n + 4)(n + 6) \equiv (2 + 2)(2 + 4)(2 + 6) \pmod{7}
$$

$$
\equiv (4)(6)(8) \pmod{7}
$$

---

### Step 2: Simplify Each Term Modulo 7

Now simplify each term modulo 7:

- $ 4 \mod 7 = 4 $
- $ 6 \mod 7 = 6 $
- $ 8 \mod 7 = 1 $

So the expression becomes:

$$
(4)(6)(1) \pmod{7}
$$

---

### Step 3: Multiply and Reduce Modulo 7

First, compute:

$$
4 \times 6 = 24
$$

Then:

$$
24 \mod 7 = 3
$$

So the final result is:

$$
3 \pmod{7}
$$

---

### Final Answer

$$
\boxed{3}
$$

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 and simplify the equation. Let me start by expanding the left side.

The left side is $(x - 3)^3 + (x - 7)^3$. I remember that $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, so maybe I can apply that formula here. Let me do that for each term.

Starting with $(x - 3)^3$:

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

Wait, let me check that again. The formula is $a^3 - 3a^2b + 3ab^2 - b^3$, so substituting $a = x$ and $b = 3$:

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

Yes, that seems right. Now for $(x - 7)^3$:

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

Let me verify that again. Using the same formula with $a = x$ and $b = 7$:

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

Okay, that looks correct. So adding the two expanded terms together:

Left side becomes:

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

Combine like terms:

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

So the left side simplifies to:

$$
2x^3 - 30x^2 + 174x - 370
$$

Now, let me expand the right side, which is $(2x - 10)^3$. Again, using the same formula. Let me recall that $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Here, $a = 2x$ and $b = 10$:

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

Calculating each term:

- $(2x)^3 = 8x^3$
- $-3(2x)^2(10) = -3*(4x^2)*10 = -120x^2$
- $3(2x)(10)^2 = 3*2x*100 = 600x$
- $-(10)^3 = -1000$

So putting it all together:

$$
8x^3 - 120x^2 + 600x - 1000
$$

Therefore, the original equation becomes:

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

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

So, bringing all terms 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 the equation simplifies to:

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

Hmm, maybe I can factor out a common factor here. Let me check if all coefficients are divisible by something. Let's see:

-6, 90, -426, 630.

Looking for a common divisor. Let's check 6:

-6 is divisible by 6, 90 is divisible by 6, 426 divided by 6 is 71, and 630 divided by 6 is 105. Wait, 426 divided by 6 is 71? Let me check:

6*70 = 420, so 6*71 = 426. Yes. And 6*105 = 630. So yes, all coefficients are divisible by 6. Let me factor that out:

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

So, dividing both sides by 6:

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

Alternatively, multiplying both sides by -1 to make the leading coefficient positive:

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

So now, we have a cubic equation: $x^3 - 15x^2 + 71x - 105 = 0$. Let me try to factor this. Maybe there are rational roots. By Rational Root Theorem, possible rational roots are factors of 105 over factors of 1, so ±1, ±3, ±5, ±7, ±15, ±21, ±35, ±105.

Let me test x=1:

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

x=3:

27 - 15*9 + 71*3 -105 = 27 -135 + 213 -105. Let's 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 cubic: 1 (x^3), -15 (x^2), 71 (x), -105 (constant).

Set up synthetic division:

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

So, the cubic factors as (x - 3)(x^2 -12x +35). Now, let's factor the quadratic: x^2 -12x +35.

Looking for two numbers that multiply to 35 and add to -12. Let me think: factors of 35 are 1 & 35, 5 &7. 5 and 7 add to 12. So since the middle term is -12x, both numbers should be negative: -5 and -7.

Therefore, x^2 -12x +35 factors as (x -5)(x -7).

So, putting it all together, the cubic factors as:

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

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

Wait, but hold on. Let me check if these roots satisfy the original equation. Because sometimes when we manipulate equations, especially expanding and moving terms around, we might introduce extraneous roots or maybe lose some? Although in this case, since we just moved terms around and factored, I think they should be valid. But let me check.

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 yes, equal. Good.

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. Equal. Good.

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. Equal. Perfect.

So all three roots are valid. Therefore, the roots are 3, 5, 7.

But wait, let me think again. The original equation was $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$. When I expanded both sides, moved everything to left, and simplified, I ended up with the cubic equation which factors into (x -3)(x -5)(x -7)=0. So all three roots are valid.

But let me think if there's another way to approach this problem, maybe more clever? Because sometimes there are identities or substitutions that can make things easier. Let me try to see.

Looking at the original equation: $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$. Notice that 2x -10 is 2(x -5). Also, note that 3 and 7 are symmetric around 5, since 5 - 3 = 2 and 7 -5 = 2. So maybe there's some symmetry here. Let me set y = x -5, so that x = y +5. Let me try substituting that.

Let me let y = x -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(y +5) -10]^3 = (2y +10 -10)^3 = (2y)^3 = 8y^3

So the equation becomes:

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

Let me compute the left side:

(y + 2)^3 + (y - 2)^3. There's a formula for a^3 + b^3, but here it's (a + b)(a^2 - ab + b^2), but maybe expanding is easier.

Alternatively, recall that (a + b)^3 + (a - b)^3 = 2a^3 + 6a b^2. Wait, let me check:

Let me expand (a + b)^3 + (a - b)^3:

= [a^3 + 3a^2b + 3ab^2 + b^3] + [a^3 - 3a^2b + 3ab^2 - b^3]

Combine terms:

a^3 + a^3 = 2a^3

3a^2b - 3a^2b = 0

3ab^2 + 3ab^2 = 6ab^2

b^3 - b^3 = 0

So total is 2a^3 + 6ab^2. Therefore, in our case, a = y, b = 2. Therefore:

Left side is 2y^3 + 6y*(2)^2 = 2y^3 + 6y*4 = 2y^3 + 24y

So equation becomes:

2y^3 + 24y = 8y^3

Bring all terms to left side:

2y^3 +24y -8y^3 = 0 => -6y^3 +24y = 0 => -6y^3 +24y = 0

Factor out -6y:

-6y(y^2 - 4) = 0 => -6y(y - 2)(y + 2) = 0

Therefore, solutions are y = 0, y = 2, y = -2.

Recall that y = x -5, so:

For y = 0: x -5 = 0 => x =5

For y = 2: x -5 = 2 => x =7

For y = -2: x -5 = -2 => x =3

So the roots are x=3, x=5, x=7. Same as before. So this method is more straightforward and avoids expanding all terms. It's clever to notice the symmetry around x=5.

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

But let me check once more if there are any other roots or if I made a mistake in the algebra. Since the original equation is a cubic, there can be at most three real roots, and we found three, so that's all.

Alternatively, maybe there's a way to see that x=5 is a solution? Let me check. If x=5, then left side is (5-3)^3 + (5-7)^3 = 8 + (-8) = 0, and right side is (10 -10)^3 = 0. So yes, x=5 is a solution. Also, since the equation is symmetric around x=5, maybe x=3 and x=7 are symmetric around 5. So that's why they are roots.

Therefore, I think the roots are indeed 3, 5, 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 begin by observing a key symmetry in the equation. The terms $x - 3$ and $x - 7$ are symmetric around $x = 5$, since $5 - 3 = 2$ and $7 - 5 = 2$. This suggests a substitution that might simplify the equation.

---

### Step 1: Substitution

Let us define a new variable:

$$
y = x - 5 \quad \Rightarrow \quad x = y + 5
$$

Substitute into the original equation:

- Left side:
  $$
  (x - 3)^3 + (x - 7)^3 = (y + 2)^3 + (y - 2)^3
  $$

- Right side:
  $$
  (2x - 10)^3 = (2(y + 5) - 10)^3 = (2y)^3 = 8y^3
  $$

So the equation becomes:

$$
(y + 2)^3 + (y - 2)^3 = 8y^3
$$

---

### Step 2: Use Identity for Sum of Cubes

Recall the identity:

$$
(a + b)^3 + (a - b)^3 = 2a^3 + 6ab^2
$$

Apply this with $a = y$, $b = 2$:

$$
(y + 2)^3 + (y - 2)^3 = 2y^3 + 6y(2)^2 = 2y^3 + 24y
$$

So the equation becomes:

$$
2y^3 + 24y = 8y^3
$$

---

### Step 3: Simplify and Solve

Bring all terms to one side:

$$
2y^3 + 24y - 8y^3 = 0 \Rightarrow -6y^3 + 24y = 0
$$

Factor:

$$
-6y(y^2 - 4) = 0 \Rightarrow -6y(y - 2)(y + 2) = 0
$$

So the solutions are:

$$
y = 0, \quad y = 2, \quad y = -2
$$

Recall that $y = x - 5$, so:

- $y = 0 \Rightarrow x = 5$
- $y = 2 \Rightarrow x = 7$
- $y = -2 \Rightarrow x = 3$

---

### Step 4: Verify the Roots

We can verify that all three values satisfy the original equation:

- For $x = 3$:
  Left: $(3 - 3)^3 + (3 - 7)^3 = 0 + (-4)^3 = -64$
  Right: $(2 \cdot 3 - 10)^3 = (-4)^3 = -64$

- For $x = 5$:
  Left: $(5 - 3)^3 + (5 - 7)^3 = 8 + (-8) = 0$
  Right: $(2 \cdot 5 - 10)^3 = 0^3 = 0$

- For $x = 7$:
  Left: $(7 - 3)^3 + (7 - 7)^3 = 64 + 0 = 64$
  Right: $(2 \cdot 7 - 10)^3 = 4^3 = 64$

All three values satisfy the equation.

---

### 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 and simplify the equation. Let me start by expanding the left side.

The left side is $(x - 3)^3 + (x - 7)^3$. I remember that $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, so maybe I can apply that formula here. Let me do that for each term.

Starting with $(x - 3)^3$:

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

Wait, let me check that again. The formula is $a^3 - 3a^2b + 3ab^2 - b^3$, so substituting $a = x$ and $b = 3$:

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

Yes, that seems right. Now for $(x - 7)^3$:

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

Let me verify that again. Using the same formula with $a = x$ and $b = 7$:

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

Okay, that looks correct. So adding the two expanded terms together:

Left side becomes:

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

Combine like terms:

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

So the left side simplifies to:

$$
2x^3 - 30x^2 + 174x - 370
$$

Now, let me expand the right side, which is $(2x - 10)^3$. Again, using the same formula. Let me recall that $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$. Here, $a = 2x$ and $b = 10$:

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

Calculating each term:

- $(2x)^3 = 8x^3$
- $-3(2x)^2(10) = -3*(4x^2)*10 = -120x^2$
- $3(2x)(10)^2 = 3*2x*100 = 600x$
- $-(10)^3 = -1000$

So putting it all together:

$$
8x^3 - 120x^2 + 600x - 1000
$$

Therefore, the original equation becomes:

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

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

So, bringing all terms 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 the equation simplifies to:

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

Hmm, maybe I can factor out a common factor here. Let me check if all coefficients are divisible by something. Let's see:

-6, 90, -426, 630.

Looking for a common divisor. Let's check 6:

-6 is divisible by 6, 90 is divisible by 6, 426 divided by 6 is 71, and 630 divided by 6 is 105. Wait, 426 divided by 6 is 71? Let me check:

6*70 = 420, so 6*71 = 426. Yes. And 6*105 = 630. So yes, all coefficients are divisible by 6. Let me factor that out:

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

So, dividing both sides by 6:

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

Alternatively, multiplying both sides by -1 to make the leading coefficient positive:

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

So now, we have a cubic equation: $x^3 - 15x^2 + 71x - 105 = 0$. Let me try to factor this. Maybe there are rational roots. By Rational Root Theorem, possible rational roots are factors of 105 over factors of 1, so ±1, ±3, ±5, ±7, ±15, ±21, ±35, ±105.

Let me test x=1:

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

x=3:

27 - 15*9 + 71*3 -105 = 27 -135 + 213 -105. Let's 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 cubic: 1 (x^3), -15 (x^2), 71 (x), -105 (constant).

Set up synthetic division:

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

So, the cubic factors as (x - 3)(x^2 -12x +35). Now, let's factor the quadratic: x^2 -12x +35.

Looking for two numbers that multiply to 35 and add to -12. Let me think: factors of 35 are 1 & 35, 5 &7. 5 and 7 add to 12. So since the middle term is -12x, both numbers should be negative: -5 and -7.

Therefore, x^2 -12x +35 factors as (x -5)(x -7).

So, putting it all together, the cubic factors as:

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

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

Wait, but hold on. Let me check if these roots satisfy the original equation. Because sometimes when we manipulate equations, especially expanding and moving terms around, we might introduce extraneous roots or maybe lose some? Although in this case, since we just moved terms around and factored, I think they should be valid. But let me check.

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 yes, equal. Good.

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. Equal. Good.

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. Equal. Perfect.

So all three roots are valid. Therefore, the roots are 3, 5, 7.

But wait, let me think again. The original equation was $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$. When I expanded both sides, moved everything to left, and simplified, I ended up with the cubic equation which factors into (x -3)(x -5)(x -7)=0. So all three roots are valid.

But let me think if there's another way to approach this problem, maybe more clever? Because sometimes there are identities or substitutions that can make things easier. Let me try to see.

Looking at the original equation: $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$. Notice that 2x -10 is 2(x -5). Also, note that 3 and 7 are symmetric around 5, since 5 - 3 = 2 and 7 -5 = 2. So maybe there's some symmetry here. Let me set y = x -5, so that x = y +5. Let me try substituting that.

Let me let y = x -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(y +5) -10]^3 = (2y +10 -10)^3 = (2y)^3 = 8y^3

So the equation becomes:

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

Let me compute the left side:

(y + 2)^3 + (y - 2)^3. There's a formula for a^3 + b^3, but here it's (a + b)(a^2 - ab + b^2), but maybe expanding is easier.

Alternatively, recall that (a + b)^3 + (a - b)^3 = 2a^3 + 6a b^2. Wait, let me check:

Let me expand (a + b)^3 + (a - b)^3:

= [a^3 + 3a^2b + 3ab^2 + b^3] + [a^3 - 3a^2b + 3ab^2 - b^3]

Combine terms:

a^3 + a^3 = 2a^3

3a^2b - 3a^2b = 0

3ab^2 + 3ab^2 = 6ab^2

b^3 - b^3 = 0

So total is 2a^3 + 6ab^2. Therefore, in our case, a = y, b = 2. Therefore:

Left side is 2y^3 + 6y*(2)^2 = 2y^3 + 6y*4 = 2y^3 + 24y

So equation becomes:

2y^3 + 24y = 8y^3

Bring all terms to left side:

2y^3 +24y -8y^3 = 0 => -6y^3 +24y = 0 => -6y^3 +24y = 0

Factor out -6y:

-6y(y^2 - 4) = 0 => -6y(y - 2)(y + 2) = 0

Therefore, solutions are y = 0, y = 2, y = -2.

Recall that y = x -5, so:

For y = 0: x -5 = 0 => x =5

For y = 2: x -5 = 2 => x =7

For y = -2: x -5 = -2 => x =3

So the roots are x=3, x=5, x=7. Same as before. So this method is more straightforward and avoids expanding all terms. It's clever to notice the symmetry around x=5.

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

But let me check once more if there are any other roots or if I made a mistake in the algebra. Since the original equation is a cubic, there can be at most three real roots, and we found three, so that's all.

Alternatively, maybe there's a way to see that x=5 is a solution? Let me check. If x=5, then left side is (5-3)^3 + (5-7)^3 = 8 + (-8) = 0, and right side is (10 -10)^3 = 0. So yes, x=5 is a solution. Also, since the equation is symmetric around x=5, maybe x=3 and x=7 are symmetric around 5. So that's why they are roots.

Therefore, I think the roots are indeed 3, 5, 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 begin by observing a key symmetry in the equation. The terms $x - 3$ and $x - 7$ are symmetric around $x = 5$, since $5 - 3 = 2$ and $7 - 5 = 2$. This suggests a substitution that might simplify the equation.

---

### Step 1: Substitution

Let us define a new variable:

$$
y = x - 5 \quad \Rightarrow \quad x = y + 5
$$

Substitute into the original equation:

- Left side:
  $$
  (x - 3)^3 + (x - 7)^3 = (y + 2)^3 + (y - 2)^3
  $$

- Right side:
  $$
  (2x - 10)^3 = (2(y + 5) - 10)^3 = (2y)^3 = 8y^3
  $$

So the equation becomes:

$$
(y + 2)^3 + (y - 2)^3 = 8y^3
$$

---

### Step 2: Use Identity for Sum of Cubes

Recall the identity:

$$
(a + b)^3 + (a - b)^3 = 2a^3 + 6ab^2
$$

Apply this with $a = y$, $b = 2$:

$$
(y + 2)^3 + (y - 2)^3 = 2y^3 + 6y(2)^2 = 2y^3 + 24y
$$

So the equation becomes:

$$
2y^3 + 24y = 8y^3
$$

---

### Step 3: Simplify and Solve

Bring all terms to one side:

$$
2y^3 + 24y - 8y^3 = 0 \Rightarrow -6y^3 + 24y = 0
$$

Factor:

$$
-6y(y^2 - 4) = 0 \Rightarrow -6y(y - 2)(y + 2) = 0
$$

So the solutions are:

$$
y = 0, \quad y = 2, \quad y = -2
$$

Recall that $y = x - 5$, so:

- $y = 0 \Rightarrow x = 5$
- $y = 2 \Rightarrow x = 7$
- $y = -2 \Rightarrow x = 3$

---

### Step 4: Verify the Roots

We can verify that all three values satisfy the original equation:

- For $x = 3$:
  Left: $(3 - 3)^3 + (3 - 7)^3 = 0 + (-4)^3 = -64$
  Right: $(2 \cdot 3 - 10)^3 = (-4)^3 = -64$

- For $x = 5$:
  Left: $(5 - 3)^3 + (5 - 7)^3 = 8 + (-8) = 0$
  Right: $(2 \cdot 5 - 10)^3 = 0^3 = 0$

- For $x = 7$:
  Left: $(7 - 3)^3 + (7 - 7)^3 = 64 + 0 = 64$
  Right: $(2 \cdot 7 - 10)^3 = 4^3 = 64$

All three values satisfy the equation.

---

### 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 find the length of each side of a regular octagon that has the same perimeter as a regular hexagon with side length 16 cm. Let me start by recalling some basic properties of 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. Since it's regular, all sides are 16 cm. A hexagon has 6 sides, so its perimeter should be 6 times 16 cm. Let me calculate that first.

Perimeter of hexagon = 6 * 16 cm = 96 cm. Wait, 6*16... 6*10 is 60, 6*6 is 36, so 60+36=96. Yes, that's right. So the perimeter is 96 cm.

Now, the problem says that a regular octagon has the same perimeter. So the perimeter of the octagon is also 96 cm. Since an octagon has 8 sides, each side of the octagon would be the perimeter divided by 8. Let me check that.

Let me denote the side length of the octagon as 's'. Then, perimeter of octagon = 8 * s. And since this is equal to the perimeter of the hexagon, which is 96 cm, we can set up the equation:

8 * s = 96 cm

To find 's', divide both sides by 8:

s = 96 / 8 = 12 cm.

Wait, that seems straightforward. But let me make sure I didn't miss anything. The problem mentions an Asymptote figure, but since I can't see the figure, I need to rely on the description. The Asymptote code draws a hexagon with a label "16 cm" on a segment between points A and B. The code defines points A and B as dir(120) and dir(60), which are directions on the unit circle. Then M is the midpoint of A and B, and the label is placed at M, North. So, the segment AB is a side of the hexagon? Wait, in a regular hexagon, each side is equal, so if AB is a side, then its length is 16 cm. But in the Asymptote code, the coordinates are given in terms of dir(angles), which are points on the unit circle. However, in reality, the figure is drawn with those coordinates, but the actual side length is given as 16 cm. So maybe the Asymptote code is just a generic drawing, and the actual side length is 16 cm. Therefore, the perimeter calculation is correct as 6*16=96 cm. Therefore, the octagon's perimeter is also 96 cm, so each side is 96/8=12 cm. Therefore, the answer is 12 cm.

But let me think again. Is there any chance that the Asymptote figure might be misleading? For example, maybe the hexagon in the figure isn't regular? But the problem says "the regular hexagon shown here", so the figure is of a regular hexagon. The Asymptote code draws a hexagon with points dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. Let me try to parse that.

The dir(theta) function in Asymptote gives a point on the unit circle at angle theta degrees. So dir(360) is the same as dir(0), which is (1,0). Then dir(60) is (cos(60°), sin(60°)) which is (0.5, √3/2). Similarly, dir(120) is (-0.5, √3/2). Dir(180) is (-1, 0), dir(240) is (-0.5, -√3/2), dir(300) is (0.5, -√3/2). So the figure is connecting dir(360) which is (1,0) to B which is dir(60) which is (0.5, √3/2), then to A which is dir(120) which is (-0.5, √3/2), then to dir(180) which is (-1, 0), then to dir(240) which is (-0.5, -√3/2), then to dir(300) which is (0.5, -√3/2), and back to dir(360). So connecting those points.

So this seems to be a regular hexagon? Let me check the distances between consecutive points. For example, the distance between dir(360) (which is (1,0)) and dir(60) (which is (0.5, √3/2)). The distance between (1,0) and (0.5, √3/2) is sqrt[(1 - 0.5)^2 + (0 - √3/2)^2] = sqrt[(0.5)^2 + (√3/2)^2] = sqrt[0.25 + 0.75] = sqrt[1] = 1. So in the unit circle, each side of the hexagon is length 1. But in the problem, the side length is given as 16 cm. Therefore, the Asymptote figure is just a unit circle hexagon, but scaled up so that the side length is 16 cm. Therefore, the perimeter of the hexagon is 6*16=96 cm, as I calculated before. Therefore, the octagon's perimeter is also 96 cm, so each side is 96/8=12 cm.

Therefore, the answer is 12 cm. But let me check again. Maybe there's a trick here? For example, sometimes problems mention perimeters but maybe there's something about the figure? But the problem says "a regular octagon has the same perimeter as the regular hexagon shown here with side length 16 cm." So the hexagon has side length 16 cm, so perimeter 96 cm. Therefore, octagon has perimeter 96 cm, so each side is 96/8=12 cm.

Alternatively, maybe the Asymptote figure is not a regular hexagon? Let me check the coordinates again. The Asymptote code draws dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. Let me check if all sides are equal.

First, dir(360) to B (dir(60)): length 1 as calculated. Then B (dir(60)) to A (dir(120)): distance between dir(60) and dir(120). Let me compute that. Coordinates of dir(60) are (cos(60°), sin(60°)) = (0.5, √3/2). Coordinates of dir(120) are (cos(120°), sin(120°)) = (-0.5, √3/2). The distance between them is sqrt[(-0.5 - 0.5)^2 + (√3/2 - √3/2)^2] = sqrt[(-1)^2 + 0] = sqrt[1] = 1. Then from A (dir(120)) to dir(180): dir(180) is (-1, 0). Distance between (-0.5, √3/2) and (-1, 0): sqrt[(-1 - (-0.5))^2 + (0 - √3/2)^2] = sqrt[(-0.5)^2 + (-√3/2)^2] = sqrt[0.25 + 0.75] = sqrt[1] = 1. Then from dir(180) to dir(240): dir(240) is (-0.5, -√3/2). Distance between (-1, 0) and (-0.5, -√3/2): sqrt[(-0.5 - (-1))^2 + (-√3/2 - 0)^2] = sqrt[(0.5)^2 + (-√3/2)^2] = sqrt[0.25 + 0.75] = 1. Then from dir(240) to dir(300): dir(300) is (0.5, -√3/2). Distance between (-0.5, -√3/2) and (0.5, -√3/2): sqrt[(0.5 - (-0.5))^2 + (-√3/2 - (-√3/2))^2] = sqrt[(1)^2 + 0] = 1. Then from dir(300) back to dir(360): dir(360) is (1, 0). Distance between (0.5, -√3/2) and (1, 0): sqrt[(1 - 0.5)^2 + (0 - (-√3/2))^2] = sqrt[(0.5)^2 + (√3/2)^2] = sqrt[0.25 + 0.75] = 1. So all sides are length 1 in the unit circle. Therefore, the figure is indeed a regular hexagon. Therefore, the side length is 16 cm, so perimeter is 6*16=96 cm. Therefore, octagon's perimeter is 96 cm, so each side is 96/8=12 cm.

Therefore, the answer is 12 cm. Wait, but maybe I need to check if the Asymptote figure is drawn with some other scaling? For example, maybe the label "16 cm" is not on a side? The Asymptote code labels "16 cm" at point M, which is the midpoint of A and B. So points A and B are dir(120) and dir(60). The segment AB is a side of the hexagon? Wait, in the drawing, the hexagon is drawn as dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. So the sides are dir(360)-B, B-A, A-dir(180), dir(180)-dir(240), dir(240)-dir(300), and dir(300)-dir(360). Therefore, AB is a side of the hexagon. Therefore, the segment AB is a side of the hexagon, and the label "16 cm" is placed at the midpoint of AB, so that segment AB is 16 cm. Therefore, the side length of the hexagon is 16 cm. Therefore, the perimeter is 6*16=96 cm. Therefore, octagon's perimeter is 96 cm, so each side is 96/8=12 cm.

Therefore, I think my answer is correct. But just to make sure, let me think again. Maybe there's a misunderstanding about the number of sides? Wait, a regular octagon has 8 sides, yes. A regular hexagon has 6 sides. So if the perimeters are equal, then 6*16 = 8*s => s= (6*16)/8 = (6/8)*16 = (3/4)*16 = 12. Yep, that's the same result.

Alternatively, maybe the problem is trying to trick me into thinking about area instead of perimeter? But the problem specifically says "perimeter". So no, that's not the case.

Alternatively, maybe the Asymptote figure is not a regular hexagon? But as we saw, all sides are equal in the Asymptote code, so it is a regular hexagon. Therefore, the side length is 16 cm.

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

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

To determine the side length of a regular octagon that has the same perimeter as a regular hexagon with side length 16 cm, we begin by analyzing the perimeters of both polygons.

---

### Step 1: Calculate the Perimeter of the Hexagon

A **regular hexagon** has 6 equal sides. If each side is 16 cm, then the perimeter is:

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

---

### Step 2: Use the Perimeter to Find the Side Length of the Octagon

A **regular octagon** has 8 equal sides. Let the side length of the octagon be $ s $. Since the perimeter of the octagon is the same as that of the hexagon, we set up the equation:

$$
8s = 96
$$

Solving for $ s $:

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

---

### Step 3: Interpret the Asymptote Diagram

The Asymptote code draws a regular hexagon with side length 16 cm, as indicated by the label "16 cm" placed at the midpoint of a side. This confirms that the hexagon in the diagram is indeed regular and has side length 16 cm. Therefore, the perimeter calculation is accurate.

---

### Final Answer

The length of each side of the regular octagon is:

$$
\boxed{12}
$$

[Assistant]
Explanation:
Okay, so I need to find the length of each side of a regular octagon that has the same perimeter as a regular hexagon with side length 16 cm. Let me start by recalling some basic properties of 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. Since it's regular, all sides are 16 cm. A hexagon has 6 sides, so its perimeter should be 6 times 16 cm. Let me calculate that first.

Perimeter of hexagon = 6 * 16 cm = 96 cm. Wait, 6*16... 6*10 is 60, 6*6 is 36, so 60+36=96. Yes, that's right. So the perimeter is 96 cm.

Now, the problem says that a regular octagon has the same perimeter. So the perimeter of the octagon is also 96 cm. Since an octagon has 8 sides, each side of the octagon would be the perimeter divided by 8. Let me check that.

Let me denote the side length of the octagon as 's'. Then, perimeter of octagon = 8 * s. And since this is equal to the perimeter of the hexagon, which is 96 cm, we can set up the equation:

8 * s = 96 cm

To find 's', divide both sides by 8:

s = 96 / 8 = 12 cm.

Wait, that seems straightforward. But let me make sure I didn't miss anything. The problem mentions an Asymptote figure, but since I can't see the figure, I need to rely on the description. The Asymptote code draws a hexagon with a label "16 cm" on a segment between points A and B. The code defines points A and B as dir(120) and dir(60), which are directions on the unit circle. Then M is the midpoint of A and B, and the label is placed at M, North. So, the segment AB is a side of the hexagon? Wait, in a regular hexagon, each side is equal, so if AB is a side, then its length is 16 cm. But in the Asymptote code, the coordinates are given in terms of dir(angles), which are points on the unit circle. However, in reality, the figure is drawn with those coordinates, but the actual side length is given as 16 cm. So maybe the Asymptote code is just a generic drawing, and the actual side length is 16 cm. Therefore, the perimeter calculation is correct as 6*16=96 cm. Therefore, the octagon's perimeter is also 96 cm, so each side is 96/8=12 cm. Therefore, the answer is 12 cm.

But let me think again. Is there any chance that the Asymptote figure might be misleading? For example, maybe the hexagon in the figure isn't regular? But the problem says "the regular hexagon shown here", so the figure is of a regular hexagon. The Asymptote code draws a hexagon with points dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. Let me try to parse that.

The dir(theta) function in Asymptote gives a point on the unit circle at angle theta degrees. So dir(360) is the same as dir(0), which is (1,0). Then dir(60) is (cos(60°), sin(60°)) which is (0.5, √3/2). Similarly, dir(120) is (-0.5, √3/2). Dir(180) is (-1, 0), dir(240) is (-0.5, -√3/2), dir(300) is (0.5, -√3/2). So the figure is connecting dir(360) which is (1,0) to B which is dir(60) which is (0.5, √3/2), then to A which is dir(120) which is (-0.5, √3/2), then to dir(180) which is (-1, 0), then to dir(240) which is (-0.5, -√3/2), then to dir(300) which is (0.5, -√3/2), and back to dir(360). So connecting those points.

So this seems to be a regular hexagon? Let me check the distances between consecutive points. For example, the distance between dir(360) (which is (1,0)) and dir(60) (which is (0.5, √3/2)). The distance between (1,0) and (0.5, √3/2) is sqrt[(1 - 0.5)^2 + (0 - √3/2)^2] = sqrt[(0.5)^2 + (√3/2)^2] = sqrt[0.25 + 0.75] = sqrt[1] = 1. So in the unit circle, each side of the hexagon is length 1. But in the problem, the side length is given as 16 cm. Therefore, the Asymptote figure is just a unit circle hexagon, but scaled up so that the side length is 16 cm. Therefore, the perimeter of the hexagon is 6*16=96 cm, as I calculated before. Therefore, the octagon's perimeter is also 96 cm, so each side is 96/8=12 cm.

Therefore, the answer is 12 cm. But let me check again. Maybe there's a trick here? For example, sometimes problems mention perimeters but maybe there's something about the figure? But the problem says "a regular octagon has the same perimeter as the regular hexagon shown here with side length 16 cm." So the hexagon has side length 16 cm, so perimeter 96 cm. Therefore, octagon has perimeter 96 cm, so each side is 96/8=12 cm.

Alternatively, maybe the Asymptote figure is not a regular hexagon? Let me check the coordinates again. The Asymptote code draws dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. Let me check if all sides are equal.

First, dir(360) to B (dir(60)): length 1 as calculated. Then B (dir(60)) to A (dir(120)): distance between dir(60) and dir(120). Let me compute that. Coordinates of dir(60) are (cos(60°), sin(60°)) = (0.5, √3/2). Coordinates of dir(120) are (cos(120°), sin(120°)) = (-0.5, √3/2). The distance between them is sqrt[(-0.5 - 0.5)^2 + (√3/2 - √3/2)^2] = sqrt[(-1)^2 + 0] = sqrt[1] = 1. Then from A (dir(120)) to dir(180): dir(180) is (-1, 0). Distance between (-0.5, √3/2) and (-1, 0): sqrt[(-1 - (-0.5))^2 + (0 - √3/2)^2] = sqrt[(-0.5)^2 + (-√3/2)^2] = sqrt[0.25 + 0.75] = sqrt[1] = 1. Then from dir(180) to dir(240): dir(240) is (-0.5, -√3/2). Distance between (-1, 0) and (-0.5, -√3/2): sqrt[(-0.5 - (-1))^2 + (-√3/2 - 0)^2] = sqrt[(0.5)^2 + (-√3/2)^2] = sqrt[0.25 + 0.75] = 1. Then from dir(240) to dir(300): dir(300) is (0.5, -√3/2). Distance between (-0.5, -√3/2) and (0.5, -√3/2): sqrt[(0.5 - (-0.5))^2 + (-√3/2 - (-√3/2))^2] = sqrt[(1)^2 + 0] = 1. Then from dir(300) back to dir(360): dir(360) is (1, 0). Distance between (0.5, -√3/2) and (1, 0): sqrt[(1 - 0.5)^2 + (0 - (-√3/2))^2] = sqrt[(0.5)^2 + (√3/2)^2] = sqrt[0.25 + 0.75] = 1. So all sides are length 1 in the unit circle. Therefore, the figure is indeed a regular hexagon. Therefore, the side length is 16 cm, so perimeter is 6*16=96 cm. Therefore, octagon's perimeter is 96 cm, so each side is 96/8=12 cm.

Therefore, the answer is 12 cm. Wait, but maybe I need to check if the Asymptote figure is drawn with some other scaling? For example, maybe the label "16 cm" is not on a side? The Asymptote code labels "16 cm" at point M, which is the midpoint of A and B. So points A and B are dir(120) and dir(60). The segment AB is a side of the hexagon? Wait, in the drawing, the hexagon is drawn as dir(360)--B--A--dir(180)--dir(240)--dir(300)--cycle. So the sides are dir(360)-B, B-A, A-dir(180), dir(180)-dir(240), dir(240)-dir(300), and dir(300)-dir(360). Therefore, AB is a side of the hexagon. Therefore, the segment AB is a side of the hexagon, and the label "16 cm" is placed at the midpoint of AB, so that segment AB is 16 cm. Therefore, the side length of the hexagon is 16 cm. Therefore, the perimeter is 6*16=96 cm. Therefore, octagon's perimeter is 96 cm, so each side is 96/8=12 cm.

Therefore, I think my answer is correct. But just to make sure, let me think again. Maybe there's a misunderstanding about the number of sides? Wait, a regular octagon has 8 sides, yes. A regular hexagon has 6 sides. So if the perimeters are equal, then 6*16 = 8*s => s= (6*16)/8 = (6/8)*16 = (3/4)*16 = 12. Yep, that's the same result.

Alternatively, maybe the problem is trying to trick me into thinking about area instead of perimeter? But the problem specifically says "perimeter". So no, that's not the case.

Alternatively, maybe the Asymptote figure is not a regular hexagon? But as we saw, all sides are equal in the Asymptote code, so it is a regular hexagon. Therefore, the side length is 16 cm.

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

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

To determine the side length of a regular octagon that has the same perimeter as a regular hexagon with side length 16 cm, we begin by analyzing the perimeters of both polygons.

---

### Step 1: Calculate the Perimeter of the Hexagon

A **regular hexagon** has 6 equal sides. If each side is 16 cm, then the perimeter is:

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

---

### Step 2: Use the Perimeter to Find the Side Length of the Octagon

A **regular octagon** has 8 equal sides. Let the side length of the octagon be $ s $. Since the perimeter of the octagon is the same as that of the hexagon, we set up the equation:

$$
8s = 96
$$

Solving for $ s $:

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

---

### Step 3: Interpret the Asymptote Diagram

The Asymptote code draws a regular hexagon with side length 16 cm, as indicated by the label "16 cm" placed at the midpoint of a side. This confirms that the hexagon in the diagram is indeed regular and has side length 16 cm. Therefore, the perimeter calculation is accurate.

---

### Final Answer

The length of each side of the regular 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, I remember that when dealing with compound inequalities like this, I can solve them step by step by isolating x. The inequality is already set up with 2(x - 1) in the middle, so maybe I can start by dividing all parts of the inequality by 2 to simplify it. Let me try that.

So, dividing each part by 2:

-4 divided by 2 is -2, right? Then the middle part is 2(x - 1) divided by 2, which simplifies to just (x - 1). And then 8 divided by 2 is 4. So now the inequality becomes:

-2 < x - 1 < 4.

Okay, that seems simpler. Now I need to solve for x. To do that, I should get rid of the -1 that's being subtracted from x. How do I do that? Well, I can add 1 to all three parts of the inequality. Let me write that down.

Adding 1 to each part:

-2 + 1 < x - 1 + 1 < 4 + 1.

Calculating each part:

-2 + 1 is -1, then x - 1 + 1 simplifies to x, and 4 + 1 is 5. So now the inequality is:

-1 < x < 5.

Wait, so that means x is between -1 and 5? Let me check if I did that correctly. Let me go back through the steps.

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

Divide all parts by 2: -2 < x - 1 < 4. That seems right because dividing each part by 2 maintains the inequality direction since 2 is positive. Then adding 1 to all parts: -2 + 1 is -1, and 4 + 1 is 5. So yes, that gives -1 < x < 5. Therefore, a is -1 and b is 5. Then a + b would be -1 + 5, which is 4. Is that correct?

Wait, let me verify by plugging in some numbers. Let me check if x = 0, which is between -1 and 5. Plugging into the original inequality: 2(0 - 1) = 2(-1) = -2. So the original inequality is -4 < -2 < 8. That's true because -4 is less than -2 and -2 is less than 8. Okay, that works.

What about x = -1? Let me check the boundary. If x = -1, then 2(-1 - 1) = 2(-2) = -4. The original inequality is -4 < -4 < 8. But -4 is not greater than -4, so x = -1 is not included. Similarly, if x = 5, then 2(5 - 1) = 2(4) = 8. The original inequality is -4 < 8 < 8? No, because 8 is not less than 8. So x = 5 is also not included. Therefore, the solution is indeed -1 < x < 5. So a is -1 and b is 5, so a + b is 4. That seems right.

But let me try another number outside the interval to make sure. Let's take x = -2. Then 2(-2 - 1) = 2(-3) = -6. The original inequality is -4 < -6 < 8? No, because -6 is less than -4, so that's not true. So x = -2 is not a solution, which is consistent with our solution.

How about x = 6? Then 2(6 - 1) = 2(5) = 10. The original inequality is -4 < 10 < 8? No, because 10 is not less than 8. So that's not a solution either. So our interval seems correct.

Wait, but let me think again. When I divided by 2, I had to make sure that 2 is positive, which it is, so the direction of the inequalities doesn't change. Then adding 1 to all parts is straightforward. So I think the steps are correct.

Alternatively, maybe I can solve the inequality without dividing first. Let me try that approach to cross-verify.

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

First, maybe I can subtract 2 from all parts? Wait, no, the middle term is 2(x - 1). Alternatively, maybe I can split the compound inequality into two separate inequalities:

-4 < 2(x - 1) and 2(x - 1) < 8.

Then solve each one separately.

Starting with the first inequality: -4 < 2(x - 1).

Divide both sides by 2: -2 < x - 1.

Then add 1 to both sides: -2 + 1 < x => -1 < x.

Now the second inequality: 2(x - 1) < 8.

Divide both sides by 2: x - 1 < 4.

Add 1 to both sides: x < 5.

So combining both inequalities: -1 < x < 5. Same result as before. So that confirms that the solution is indeed -1 < x < 5. Therefore, a is -1 and b is 5, so a + b is 4.

Therefore, the answer should be 4. Wait, but let me check once more. Maybe I made a mistake in arithmetic?

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

Divide by 2: -2 < x - 1 < 4. Then add 1: -1 < x < 5. Yes. So a is -1, b is 5. Sum is 4. Seems straightforward.

Alternatively, maybe I can think of it as solving for x in the middle. Let me try another way. Let me expand the original inequality.

Original: -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 simplifies to: -2 < 2x < 10.

Then divide all parts by 2: -1 < x < 5. Same result again. So that's three different methods leading to the same answer. Therefore, I can be confident that the solution is correct.

Therefore, a is -1 and b is 5, so a + b is -1 + 5 = 4. Therefore, the answer is 4.

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

To solve the inequality:

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

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

---

### Step 1: Simplify the inequality

We begin by dividing all parts of the inequality by 2, since 2 is a positive number and does not change the direction of the inequality:

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

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

---

### Step 2: Solve for $ x $

Now, we isolate $ 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 $

From the simplified inequality $ -1 < x < 5 $, we identify:

- $ a = -1 $
- $ b = 5 $

---

### Step 4: Compute $ a + b $

$$
a + b = -1 + 5 = 4
$$

---

### 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, I remember that when dealing with compound inequalities like this, I can solve them step by step by isolating x. The inequality is already set up with 2(x - 1) in the middle, so maybe I can start by dividing all parts of the inequality by 2 to simplify it. Let me try that.

So, dividing each part by 2:

-4 divided by 2 is -2, right? Then the middle part is 2(x - 1) divided by 2, which simplifies to just (x - 1). And then 8 divided by 2 is 4. So now the inequality becomes:

-2 < x - 1 < 4.

Okay, that seems simpler. Now I need to solve for x. To do that, I should get rid of the -1 that's being subtracted from x. How do I do that? Well, I can add 1 to all three parts of the inequality. Let me write that down.

Adding 1 to each part:

-2 + 1 < x - 1 + 1 < 4 + 1.

Calculating each part:

-2 + 1 is -1, then x - 1 + 1 simplifies to x, and 4 + 1 is 5. So now the inequality is:

-1 < x < 5.

Wait, so that means x is between -1 and 5? Let me check if I did that correctly. Let me go back through the steps.

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

Divide all parts by 2: -2 < x - 1 < 4. That seems right because dividing each part by 2 maintains the inequality direction since 2 is positive. Then adding 1 to all parts: -2 + 1 is -1, and 4 + 1 is 5. So yes, that gives -1 < x < 5. Therefore, a is -1 and b is 5. Then a + b would be -1 + 5, which is 4. Is that correct?

Wait, let me verify by plugging in some numbers. Let me check if x = 0, which is between -1 and 5. Plugging into the original inequality: 2(0 - 1) = 2(-1) = -2. So the original inequality is -4 < -2 < 8. That's true because -4 is less than -2 and -2 is less than 8. Okay, that works.

What about x = -1? Let me check the boundary. If x = -1, then 2(-1 - 1) = 2(-2) = -4. The original inequality is -4 < -4 < 8. But -4 is not greater than -4, so x = -1 is not included. Similarly, if x = 5, then 2(5 - 1) = 2(4) = 8. The original inequality is -4 < 8 < 8? No, because 8 is not less than 8. So x = 5 is also not included. Therefore, the solution is indeed -1 < x < 5. So a is -1 and b is 5, so a + b is 4. That seems right.

But let me try another number outside the interval to make sure. Let's take x = -2. Then 2(-2 - 1) = 2(-3) = -6. The original inequality is -4 < -6 < 8? No, because -6 is less than -4, so that's not true. So x = -2 is not a solution, which is consistent with our solution.

How about x = 6? Then 2(6 - 1) = 2(5) = 10. The original inequality is -4 < 10 < 8? No, because 10 is not less than 8. So that's not a solution either. So our interval seems correct.

Wait, but let me think again. When I divided by 2, I had to make sure that 2 is positive, which it is, so the direction of the inequalities doesn't change. Then adding 1 to all parts is straightforward. So I think the steps are correct.

Alternatively, maybe I can solve the inequality without dividing first. Let me try that approach to cross-verify.

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

First, maybe I can subtract 2 from all parts? Wait, no, the middle term is 2(x - 1). Alternatively, maybe I can split the compound inequality into two separate inequalities:

-4 < 2(x - 1) and 2(x - 1) < 8.

Then solve each one separately.

Starting with the first inequality: -4 < 2(x - 1).

Divide both sides by 2: -2 < x - 1.

Then add 1 to both sides: -2 + 1 < x => -1 < x.

Now the second inequality: 2(x - 1) < 8.

Divide both sides by 2: x - 1 < 4.

Add 1 to both sides: x < 5.

So combining both inequalities: -1 < x < 5. Same result as before. So that confirms that the solution is indeed -1 < x < 5. Therefore, a is -1 and b is 5, so a + b is 4.

Therefore, the answer should be 4. Wait, but let me check once more. Maybe I made a mistake in arithmetic?

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

Divide by 2: -2 < x - 1 < 4. Then add 1: -1 < x < 5. Yes. So a is -1, b is 5. Sum is 4. Seems straightforward.

Alternatively, maybe I can think of it as solving for x in the middle. Let me try another way. Let me expand the original inequality.

Original: -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 simplifies to: -2 < 2x < 10.

Then divide all parts by 2: -1 < x < 5. Same result again. So that's three different methods leading to the same answer. Therefore, I can be confident that the solution is correct.

Therefore, a is -1 and b is 5, so a + b is -1 + 5 = 4. Therefore, the answer is 4.

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

To solve the inequality:

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

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

---

### Step 1: Simplify the inequality

We begin by dividing all parts of the inequality by 2, since 2 is a positive number and does not change the direction of the inequality:

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

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

---

### Step 2: Solve for $ x $

Now, we isolate $ 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 $

From the simplified inequality $ -1 < x < 5 $, we identify:

- $ a = -1 $
- $ b = 5 $

---

### Step 4: Compute $ a + b $

$$
a + b = -1 + 5 = 4
$$

---

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