To solve the equation \((x - 1)(x^2 - 9) = -x - 3\), follow these steps:

1. Expand the left-hand side of the equation:
\[
(x - 1)(x^2 - 9) = x(x^2 - 9) - 1(x^2 - 9) = x^3 - 9x - x^2 + 9
\]
So, the equation becomes:
\[
x^3 - x^2 - 9x + 9 = -x - 3
\]

2. Move all terms to one side of the equation to set it to zero:
\[
x^3 - x^2 - 9x + 9 + x + 3 = 0
\]
Simplify the equation:
\[
x^3 - x^2 - 8x + 12 = 0
\]

3. Now, we need to find the roots of the polynomial \(x^3 - x^2 - 8x + 12 = 0\). We can use the Rational Root Theorem to test possible rational roots. The possible rational roots are the factors of the constant term (12) divided by the factors of the leading coefficient (1), which are:
\[
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12
\]

4. Test these possible roots by substitution:

- For \(x = 1\):
\[
1^3 - 1^2 - 8(1) + 12 = 1 - 1 - 8 + 12 = 4 \quad (\text{not a root})
\]

- For \(x = -1\):
\[
(-1)^3 - (-1)^2 - 8(-1) + 12 = -1 - 1 + 8 + 12 = 18 \quad (\text{not a root})
\]

- For \(x = 2\):
\[
2^3 - 2^2 - 8(2) + 12 = 8 - 4 - 16 + 12 = 0 \quad (\text{a root})
\]

Since \(x = 2\) is a root, we can factor \(x - 2\) out of the polynomial \(x^3 - x^2 - 8x + 12\).

5. Perform polynomial division or use synthetic division to factor \(x^3 - x^2 - 8x + 12\) by \(x - 2\):

\[
x^3 - x^2 - 8x + 12 = (x - 2)(x^2 + x - 6)
\]

6. Factor the quadratic \(x^2 + x - 6\):
\[
x^2 + x - 6 = (x + 3)(x - 2)
\]

So, the complete factorization is:
\[
x^3 - x^2 - 8x + 12 = (x - 2)^2(x + 3)
\]

7. Set each factor to zero to find the roots:
\[
(x - 2)^2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]

Therefore, the solutions to the equation \((x - 1)(x^2 - 9) = -x - 3\) are:
\[
x = 2 \quad \text{and} \quad x = -3
\]