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You've provided a fraction involving a polynomial:

\[
\frac{x^3 - 1}{(x + 1)(x - 2)}
\]

To decompose this into partial fractions, we first factor the numerator, \(x^3 - 1\).

### Step 1: Factor the Numerator

The numerator \(x^3 - 1\) can be factored using the difference of cubes formula:

\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]

So the expression becomes:

\[
\frac{(x - 1)(x^2 + x + 1)}{(x + 1)(x - 2)}
\]

### Step 2: Set Up Partial Fractions

We express the fraction as a sum of partial fractions:

\[
\frac{(x - 1)(x^2 + x + 1)}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}
\]

### Step 3: Multiply by the Denominator

Multiply both sides by \((x + 1)(x - 2)\) to get:

\[
(x - 1)(x^2 + x + 1) = A(x - 2) + B(x + 1)
\]

### Step 4: Expand and Collect Like Terms

Expand both sides:

\[
x^3 - 1 = A(x - 2) + B(x + 1)
\]

Now, expand \(A(x - 2)\) and \(B(x + 1)\):

\[
x^3 - 1 = A(x - 2) + B(x + 1)
\]

To find \(A\) and \(B\), equate the coefficients of corresponding powers of \(x\):

1. Coefficient of \(x^2\): \(A + B = 0\)
2. Coefficient of \(x\): \(A \times (-2) + B \times 1 = 1\)

Solve this system to find the values of \(A\) and \(B\). 

Let me know if you want to proceed with the solution or need further explanations.

Would you like details or have any questions?

Here are some related questions:
1. How do you solve the system of equations for \(A\) and \(B\)?
2. Can you explain how to factor the difference of cubes?
3. What are partial fractions used for in calculus?
4. Can you provide an example of a more complex partial fraction decomposition?
5. How do you integrate functions with partial fractions?
6. What are the special cases when partial fractions cannot be applied directly?
7. Can you explain how to handle repeated roots in partial fractions?
8. How do you verify the partial fraction decomposition?

**Tip:** Always check your final answer by substituting the partial fractions back into the original expression.

Learn how to decompose a polynomial fraction into partial fractions step-by-step. This problem involves the difference of cubes formula and factorization techniques. Suitable for college-level students and those studying advanced calculus.

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