Okay, so I need to find the ordered pair (a, b) where a is not zero and b is a real number such that when you take f(x) minus f(a), it's 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 me write down the equation they provided: f(x) - f(a) = (x - a)²(x - b). So, if I expand both sides, they should be equal for all x, which means the coefficients of the corresponding powers of x should match. That makes sense because polynomials are equal if and only if their coefficients are equal.

Let me compute the left-hand side (LHS) first. LHS is f(x) - f(a). Since f(x) is x³ + 3x² + 1, subtracting f(a) would give me x³ + 3x² + 1 - (a³ + 3a² + 1). Simplifying that, the 1's cancel out, so LHS becomes x³ + 3x² - a³ - 3a². Alternatively, I can write that as x³ - a³ + 3x² - 3a². 

Now, the right-hand side (RHS) is (x - a)²(x - b). Let me expand that. First, (x - a)² is x² - 2ax + a². Then multiplying by (x - b) gives:

(x² - 2ax + a²)(x - b) = x³ - bx² - 2ax² + 2abx + a²x - a²b.

Combine like terms:

x³ + (-b - 2a)x² + (2ab + a²)x - a²b.

So, RHS is x³ + (-b - 2a)x² + (2ab + a²)x - a²b.

Now, setting LHS equal to RHS:

LHS: x³ + 3x² - a³ - 3a²

RHS: x³ + (-b - 2a)x² + (2ab + a²)x - a²b

Therefore, equate the coefficients of corresponding powers of x:

1. Coefficient of x³: Both sides have 1. So, nothing to do here.

2. Coefficient of x²: LHS is 3, RHS is (-b - 2a). Therefore, 3 = -b - 2a. Let me write that as equation (1): -b - 2a = 3.

3. Coefficient of x: LHS has 0 (since there's no x term), RHS has (2ab + a²). Therefore, 0 = 2ab + a². Let's call this equation (2): 2ab + a² = 0.

4. Constant term: LHS is -a³ - 3a², RHS is -a²b. Therefore, -a³ - 3a² = -a²b. Let's write this as equation (3): -a³ - 3a² = -a²b. Multiply both sides by -1: a³ + 3a² = a²b. Then, divide both sides by a² (since a ≠ 0, so division is allowed): a + 3 = b. So equation (3) simplifies to b = a + 3.

Now, we have three equations:

1. -b - 2a = 3

2. 2ab + a² = 0

3. b = a + 3

Let me use equation (3) to substitute b in equations (1) and (2).

Substituting b = a + 3 into equation (1):

- (a + 3) - 2a = 3

Simplify:

- a - 3 - 2a = 3

Combine like terms:

-3a - 3 = 3

Add 3 to both sides:

-3a = 6

Divide by -3:

a = -2

Now, substitute a = -2 into equation (3): b = (-2) + 3 = 1

So, a = -2, b = 1. Let me check if these satisfy equation (2).

Equation (2): 2ab + a² = 0

Plugging a = -2 and b = 1:

2*(-2)(1) + (-2)² = (-4) + 4 = 0. Yes, that works.

Therefore, the ordered pair is (-2, 1). Let me verify this by substituting back into the original equation.

Compute f(x) - f(a):

f(x) = x³ + 3x² + 1

f(a) = (-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 and b = 1:

(x - (-2))²(x - 1) = (x + 2)²(x - 1)

First, expand (x + 2)²: x² + 4x + 4

Multiply by (x - 1):

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

Combine like terms:

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

Which matches f(x) - f(a). So, the solution is correct.

Therefore, the ordered pair is (-2, 1).

**Final Answer**
The ordered pair is \boxed{(-2, 1)}.
Therefore, the ordered pair $(a, b)$ is $\boxed{(-2, 1)}$.
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