<Question>
Find the roots of $(x - 3)^3 + (x -7)^3 = (2x - 10)^3.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the roots of the equation \((x - 3)^3 + (x - 7)^3 = (2x - 10)^3\). Hmm, let's start by understanding what this equation is saying. On the left side, we have the sum of two cubes: \((x - 3)^3\) and \((x - 7)^3\). On the right side, there's a single cube term: \((2x - 10)^3\). 

First, I notice that \(2x - 10\) is actually twice the average of \(x - 3\) and \(x - 7\). Let me check that.
The average of \(x - 3\) and \(x - 7\) would be \(\frac{(x - 3) + (x - 7)}{2} = \frac{2x - 10}{2} = x - 5\). So, \(2x - 10\) is indeed twice that average.

This makes me think of the identity for the sum of cubes. The standard identity is \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). But here, the right side is \((a + b)^3\), where \(a = x - 3\) and \(b = x - 7\). Let me expand both sides to see if they can be simplified.

<prompt_1.1> Expand both sides of the equation using the identity for the sum of cubes and compare the results. </prompt_1.1>
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Wait, but maybe there's another approach here. Since the equation is \(a^3 + b^3 = (a + b)^3\), where \(a = x - 3\) and \(b = x - 7\), perhaps we can use the identity that \(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\). Let me substitute that in.

<prompt_1.2> Use the identity \(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\) to solve the equation. </prompt_1.2>
-----

Another way to approach this might be to substitute variables to simplify the expression. Let me try setting \(y = x - 5\).

<prompt_1.3> Use substitution to simplify the equation and find the roots. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's compute both sides step by step. 

First, let's expand the left side:

\((x - 3)^3 + (x - 7)^3\).

Expanding each cube individually:

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

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

Adding these together:

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

Combine like terms:

\(2x^3 - 30x^2 + 174x - 370\)

Now, the right side is \((2x - 10)^3\). Let's expand that:

First, note that \((a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3\). Here, \(a = 2x\), \(b = 10\):

\((2x)^3 - 3*(2x)^2*10 + 3*(2x)*(10)^2 - 10^3\)

Calculate each term:

\(8x^3 - 3*4x^2*10 + 3*2x*100 - 1000\)

Simplify:

\(8x^3 - 120x^2 + 600x - 1000\)

So now, the equation becomes:

Left side: \(2x^3 - 30x^2 + 174x - 370\)

Right side: \(8x^3 - 120x^2 + 600x - 1000\)

Subtract the right side from both sides to set the equation to zero:

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

Combine like terms:

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

Calculates to:

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

Hmm, so we have a cubic equation: \(-6x^3 + 90x^2 - 426x + 630 = 0\). Let me check my calculations again to make sure I didn't make a mistake. Let's verify the subtraction step:

Left side terms: 2x³, -30x², 174x, -370

Minus right side terms: -8x³, +120x², -600x, +1000

So adding the left side terms and the negative of the right side terms:

2x³ -8x³ = -6x³

-30x² +120x² = 90x²

174x -600x = -426x

-370 +1000 = 630

Yes, that seems correct. So the equation is -6x³ +90x² -426x +630 =0.
</execution_1.1>

<execution_1.2>
So, substituting into the original equation:

\((a + b)^3 - 3ab(a + b) = (a + b)^3\)

Subtracting \((a + b)^3\) from both sides gives:

\(-3ab(a + b) = 0\)

So, this equation is satisfied if either \(a = 0\), \(b = 0\), or \(a + b = 0\).
</execution_1.2>

<execution_1.3>
Then, \(x = y + 5\). Let's substitute this into the original equation:

\(( (y + 5) - 3 )^3 + ( (y + 5) - 7 )^3 = (2(y + 5) - 10)^3\)

Simplify each term:

\((y + 2)^3 + (y - 2)^3 = (2y + 10 - 10)^3\)

Which simplifies to:

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

Expanding the left side:

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

Combine like terms:

\(2y^3 + 24y = 8y^3\)

Bring all terms to one side:

\(2y^3 + 24y - 8y^3 = 0\)

Simplify:

\(-6y^3 + 24y = 0\)

Factor out common terms:

\(-6y(y^2 - 4) = 0\)

Which gives:

\(-6y(y - 2)(y + 2) = 0\)

So, the solutions are \(y = 0\), \(y = 2\), and \(y = -2\). Converting back to \(x\), since \(x = y + 5\):

For \(y = 0\): \(x = 5\)
For \(y = 2\): \(x = 7\)
For \(y = y = -2\): \(x = 3\)

Same solutions as before. So, this substitution method also leads us to the roots 3, 5, 7.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me factor out a common factor first. All coefficients are even except for the constant term.

<prompt_2.1> Factor out a common factor from the cubic equation and simplify it. </prompt_2.1>
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Based on execution_1.2:
So either \(a = 0\), which would mean \(x - 3 = 0\) leading to \(x = 3\); or \(b = 0\), which would mean \(x - 7 = 0\) leading to \(x = 7\); or \(a + b = 0\), which is \((x - 3) + (x - 7) = 0\). Let's solve that: \(2x - 10 = 0\) leads to \(x = 5\).
-----

Based on execution_1.1:
Let's try to factor this cubic equation. Maybe there's a common factor or rational roots.

<prompt_2.2> Factor the cubic equation using the Rational Root Theorem and verify the factorization. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's factor out -6:

\(-6( x^3 - 15x^2 + 71x - 65 ) = 0\)

Since the product is zero, we can ignore the -6 and focus on solving:

\(x^3 - 15x^2 + 71x - 65 = 0\)
</execution_2.1>

<execution_2.2>
First, let's factor out a -6:

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

So, the equation simplifies to \(x^3 - 15x^2 + 71x - 105 = 0\).

Now, let's look for rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (105) divided by factors of the leading coefficient (1). So possible roots are ±1, ±3, ±5, ±7, ±15, ±21, ±35, ±105.

Let's test these candidates. Let's start with x=3:

\(3^3 -15*(3)^2 +71*3 -105 = 27 - 135 + 213 -105 = (27 -135) + (213 -105) = (-108) + 108 = 0\). So, x=3 is a root.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
Therefore, (x - 3) is a factor. Let's perform polynomial division or use synthetic division to factor the cubic.

Using synthetic division with x=3:

<prompt_3.1> Perform synthetic division on the cubic equation using the root found. </prompt_3.1>
-----

Based on execution_2.2:
Wait, but earlier when we factored the original equation using the identity, we found roots at 3,5,7. Now, according to this, we have a root at x=3, and then a quadratic.
Let's see what the quadratic is.

After factoring out (x - 3), we have:

\(x^2 - 12x + 35 = 0\)

<prompt_3.2> Solve the quadratic equation obtained from the cubic factorization. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Coefficients of the cubic: 1 (x³), -15 (x²), 71 (x), -105 (constant)

Bring down the 1.

Multiply 1 by 3: 3. Add to -15: -12.

Multiply -12 by 3: -36. Add to 71: 35.

Multiply 35 by 3: 105. Add to -105: 0. Perfect, no remainder.

So, the cubic factors into (x - 3)(x² - 12x + 35) = 0.

Now, factor the quadratic: x² -12x +35. Let's look for two numbers that multiply to 35 and add to -12. Those numbers are -5 and -7.

So, the quadratic factors as (x - 5)(x - 7). Therefore, the cubic equation factors into:

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

Thus, the roots are x = 3, x = 5, and x = 7.
</execution_3.1>

<execution_3.2>
Factoring this quadratic: (x - 5)(x - 7) = 0, so roots at x=5 and x=7.

Wait, so the cubic factors into (x - 3)(x - 5)(x - 7) = 0, giving roots at 3,5,7. But hold on, the original equation was \((x - 3)^3 + (x - 7)^3 = (2x - 10)^3\). So, if we substitute x=5 into the original equation, let's check:

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

Right side: (10 - 10)^3 = 0^3 = 0. So, that works.

Similarly, x=3:

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

Right side: (6 -10)^3 = (-4)^3 = -64. So, that works.

x=7:

Left side: 4^3 + 0^3 = 64 + 0 = 64

Right side: (14 -10)^3 = 4^3 = 64. That works too.

So all three roots are valid.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait a minute, but let me check if these are indeed the solutions. Let's substitute each back into the original equation to verify.

<prompt_4.1> Verify the roots by substituting them back into the original equation. </prompt_4.1>
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Based on execution_3.2:
But wait, the original equation is a cubic equation, which can have up to three real roots. But when we first approached it by expanding both sides, we ended up with a cubic equation which factors into (x - 3)(x - 5)(x - 7) = 0, so indeed the roots are 3,5,7.
But let me think again—is there a possibility of any other roots? For example, when we used the identity \(a^3 + b^3 = (a + b)^3\) leading to \(a=0\), \(b=0\), or \(a + b = 0\), which gives exactly these three roots. So, in this case, all roots are real and there are three of them.
-----

Based on execution_1.1, execution_2.2, execution_3.1:
But let me just recap to ensure I didn't miss anything. The original equation simplifies to a cubic equation which factors into three linear terms, each corresponding to one of the roots we found. All three roots satisfy the original equation when substituted back in.
Therefore, the roots are x=3, x=5, and x=7.

**Final Answer**
The roots are \boxed{3}, \boxed{5}, and \boxed{7}.
</Plan_4>

<Execution_4>
<execution_4.1>
First, x = 3:

Original equation: \((3 - 3)^3 + (3 - 7)^3 = (2*3 - 10)^3\)

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

Right side: \((-4)^3 = -64\). Check.
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the roots of the equation $(x - 3)^3 + (x - 7)^3 = (2x - 10)^3$ are $\boxed{3}$, $\boxed{5}$, and $\boxed{7}$.
</Final_answer>
</think>