<Question>
Let $f(x) = x^3 + 3x^2 + 1.$  There exist real numbers $a \neq 0$ and $b,$ such that
\[f(x) - f(a) = (x - a)^2 (x - b).\]Enter the ordered pair $(a,b).$
</Question>

<think>
<Plan_1>
Okay, so I need to find real numbers a (not zero) and b such that when I subtract f(a) from f(x), the result is equal to (x - a)^2 times (x - b). The function f(x) is given as x³ + 3x² + 1. Hmm, let me start by understanding what the problem is asking.

First, let's write down the equation they gave:

f(x) - f(a) = (x - a)²(x - b)

So, if I subtract f(a) from f(x), the left-hand side (LHS) becomes a cubic polynomial minus a constant (since f(a) is a constant). The right-hand side (RHS) is a cubic polynomial as well, since (x - a)² is quadratic and (x - b) is linear, so multiplying them gives a cubic. That makes sense because both sides are cubic polynomials.

Since these two polynomials are equal for all x, their coefficients must be equal. So, maybe I can expand both sides and then equate the coefficients of corresponding powers of x. That should give me a system of equations to solve for a and b.

Let me start by computing f(x) - f(a).

<prompt_1.1> Compute f(x) - f(a) by expanding and simplifying the expression. </prompt_1.1>

Now, let's expand the right-hand side: (x - a)²(x - b).

First, expand (x - a)²:

(x - a)² = x² - 2a x + a²

Now multiply this by (x - b):

(x² - 2a x + a²)(x - b) = x³ - b x² - 2a x² + 2a b x + a² x - a² b

Combine like terms:

x³ + (-b - 2a) x² + (2a b + a²) x - a² b

So, the expanded form is:

x³ + (-b - 2a) x² + (2a b + a²) x - a² b

Now, set this equal to the left-hand side, which is f(x) - f(a):

<prompt_1.2> Expand the right-hand side and equate the coefficients of the expanded forms of both sides. </prompt_1.2>
</Plan_1>

<Execution_1>
<execution_1.1>
f(x) = x³ + 3x² + 1

f(a) = a³ + 3a² + 1

Therefore, f(x) - f(a) = (x³ + 3x² + 1) - (a³ + 3a² + 1) = x³ - a³ + 3x² - 3a²

Simplify the terms:

x³ - a³ can be factored as (x - a)(x² + a x + a²)

Similarly, 3x² - 3a² = 3(x² - a²) = 3(x - a)(x + a)

So, factoring out (x - a) from both terms:

f(x) - f(a) = (x - a)[x² + a x + a² + 3(x + a)]

Let me expand the expression inside the brackets:

x² + a x + a² + 3x + 3a = x² + (a + 3) x + (a² + 3a)

Therefore, f(x) - f(a) = (x - a)(x² + (a + 3)x + a² + 3a)
</execution_1.1>

<execution_1.2>
x³ + 3x² + 1 - [a³ + 3a² + 1] = x³ + 3x² + 1 - a³ - 3a² - 1 = x³ + 3x² - a³ - 3a²

Wait, hold on. Let me compute f(x) - f(a) properly.

f(x) = x³ + 3x² + 1

f(a) = a³ + 3a² + 1

Therefore, f(x) - f(a) = x³ + 3x² + 1 - (a³ + 3a² + 1) = x³ + 3x² - a³ - 3a²
</execution_1.2>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
So, LHS is x³ + 3x² - a³ - 3a²

RHS is x³ + (-b - 2a)x² + (2a b + a²)x - a² b

Since these polynomials are equal for all x, their coefficients must be equal. Therefore, we can set up equations by equating the coefficients of x³, x², x, and the constant term.

<prompt_2.1> Set up equations by equating the coefficients of the expanded forms of both sides. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Let's write those equations:

1. Coefficient of x³: On LHS it's 1, on RHS it's 1. So, 1 = 1. That's always true, so no information here.

2. Coefficient of x²: LHS is 3, RHS is (-b - 2a). Therefore:

3 = -b - 2a  … (Equation 1)

3. Coefficient of x: LHS has no x term, which means the coefficient is 0. RHS has (2a b + a²). Therefore:

0 = 2a b + a²  … (Equation 2)

4. Constant term: LHS has -a³ - 3a² (since f(x) - f(a) = x³ + 3x² - a³ - 3a²), and the RHS has -a² b. Therefore:

-a³ - 3a² = -a² b  … (Equation 3)

So, now we have three equations:

Equation 1: 3 = -b - 2a

Equation 2: 0 = 2a b + a²

Equation 3: -a³ - 3a² = -a² b
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Our variables are a and b. Let's try to solve these equations.

First, from Equation 1: 3 = -b - 2a. Let's solve for b:

<prompt_3.1> Solve Equation 1 for b and substitute into the other equations. </prompt_3.1>
-----

From Equation 3: -a³ - 3a² = -a² b

Divide both sides by -a² (since a ≠ 0, we can divide):

<prompt_3.2> Verify and use Equation 3 to find an expression for b. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
=> -b = 3 + 2a

Multiply both sides by -1:

b = -3 - 2a  … (Equation 1a)

So, we can express b in terms of a. Let's substitute this into Equations 2 and 3.

Equation 2: 0 = 2a b + a². Substitute b from Equation 1a:

0 = 2a(-3 - 2a) + a²

Compute 2a(-3 - 2a):

= -6a - 4a²

Therefore:

0 = -6a - 4a² + a²

Combine like terms:

0 = -6a - 3a²

Factor out -3a:

0 = -3a(2 + a)

So, either -3a = 0 => a = 0, or (2 + a) = 0 => a = -2

But the problem states that a ≠ 0, so we must have a = -2.
</execution_3.1>

<execution_3.2>
(a³ + 3a²)/a² = b

Simplify:

(a + 3) = b

So, Equation 3 gives us b = a + 3
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Now, substitute a = -2 into Equation 1a to find b:

<prompt_4.1> Substitute a = -2 into Equation 1a to find b. </prompt_4.1>
-----

Based on execution_3.2:
Wait, but from Equation 1a, we have b = -3 -2a. Let's see if these two expressions for b agree.

<prompt_4.2> Verify if the expressions for b from Equation 1a and Equation 3 are consistent. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
b = -3 - 2*(-2) = -3 + 4 = 1

So, a = -2, b = 1. Let's check if these values satisfy all three equations.

Check Equation 1: 3 = -b -2a => 3 = -1 -2*(-2) => -1 + 4 = 3. Correct.

Equation 2: 0 = 2a b + a² => 0 = 2*(-2)(1) + (-2)² => -4 + 4 = 0. Correct.

Equation 3: Let's check if with a = -2, does -a³ -3a² equal -a² b?

Left side: -(-2)^3 -3*(-2)^2 = -(-8) -3*(4) = 8 -12 = -4

Right side: -(-2)^2 * b = -4 *1 = -4. Correct.

So, all equations are satisfied. Therefore, the ordered pair is (-2, 1).
</execution_4.1>

<execution_4.2>
From Equation 3: b = a + 3

From Equation 1a: b = -3 -2a

Set them equal:

a + 3 = -3 -2a

Solve for a:

a + 2a = -3 -3

3a = -6

a = -2

Which is consistent with what we found earlier. So, there's no conflict here.
</execution_4.2>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, but let me verify this by plugging back into the original equation to make sure.

<prompt_5.1> Verify the solution by plugging a = -2 and b = 1 back into the original equation. </prompt_5.1>
-----

Based on execution_4.2:
Therefore, the answer is (-2,1).
**Final Answer**
The ordered pair is \boxed{(-2, 1)}.
</Plan_5>

<Execution_5>
<execution_5.1>
Given a = -2, b =1.

Compute f(x) - f(a):

f(x) = x³ +3x² +1

f(a) = f(-2) = (-2)^3 +3*(-2)^2 +1 = -8 + 12 +1 = 5

So, f(x) - f(a) = x³ +3x² +1 -5 = x³ +3x² -4

Now, compute (x - a)^2 (x - b) with a = -2, b =1:

(x - (-2))² (x -1) = (x +2)^2 (x -1)

First compute (x +2)^2 = x² +4x +4

Multiply by (x -1):

(x² +4x +4)(x -1) = x³ -x² +4x² -4x +4x -4 = x³ +3x² -4

Which is equal to f(x) - f(a). So, it checks out.

Therefore, the solution is a = -2, b =1, so the ordered pair is (-2,1).
</execution_5.1>
</Execution_5>

<Final_answer>
Therefore, the ordered pair $(a,b)$ is $\boxed{(-2,1)}$.
</Final_answer>
</think>