To decompose the rational function $\frac{x^3 - 2}{(x + 1)^2(x - 1)}$ into partial fractions, we follow these steps:

Step 1: Set up the form of the partial fraction decomposition

Since the denominator $(x + 1)^2(x - 1)$ consists of linear factors (some repeated), the partial fraction decomposition should have the following form:

$\frac{x^3 - 2}{(x + 1)^2(x - 1)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 1}$

Step 2: Multiply through by the denominator

To eliminate the fractions, multiply both sides by $(x + 1)^2(x - 1)$:

$x^3 - 2 = A(x + 1)(x - 1) + B(x - 1) + C(x + 1)^2$

Step 3: Expand and collect like terms

Expand each term on the right-hand side:

1. $A(x + 1)(x - 1)$ simplifies to $A(x^2 - 1)$.
2. $B(x - 1)$ stays as $B(x - 1)$.
3. $C(x + 1)^2$ expands to $C(x^2 + 2x + 1)$.

So, we have:

$x^3 - 2 = A(x^2 - 1) + B(x - 1) + C(x^2 + 2x + 1)$

Now expand and collect the terms:

$x^3 - 2 = (A + C)x^2 + (B + 2C)x + (-A - B + C)$

Step 4: Equate coefficients

Now, compare the coefficients of $x^3$, $x^2$, $x$, and the constant term on both sides of the equation:

• For $x^2$: $A + C = 0$
• For $x$: $B + 2C = 0$
• For the constant term: $-A - B + C = -2$

Step 5: Solve the system of equations

Now, solve the system of linear equations:

1. $A + C = 0 \implies A = -C$
2. $B + 2C = 0 \implies B = -2C$
3. $-A - B + C = -2$

Substitute $A = -C$ and $B = -2C$ into the third equation:

$-(-C) - (-2C) + C = -2$

Simplify:

$C + 2C + C = -2 \implies 4C = -2 \implies C = -\frac{1}{2}$

Now substitute $C = -\frac{1}{2}$ into the other equations:

$A = -C = \frac{1}{2}, \quad B = -2C = 1$

Step 6: Write the final partial fraction decomposition

Substitute $A$, $B$, and $C$ back into the partial fraction form:

$\frac{x^3 - 2}{(x + 1)^2(x - 1)} = \frac{\frac{1}{2}}{x + 1} + \frac{1}{(x + 1)^2} + \frac{-\frac{1}{2}}{x - 1}$

This simplifies to:

$\frac{x^3 - 2}{(x + 1)^2(x - 1)} = \frac{1}{2(x + 1)} + \frac{1}{(x + 1)^2} - \frac{1}{2(x - 1)}$

This is the partial fraction decomposition.

Would you like more details, or do you have any questions?

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Here are 5 related questions you might find interesting:

1. How would you decompose a fraction where the numerator has a higher degree than the denominator?
2. What happens if one of the roots in the denominator is a repeated quadratic factor?
3. Can partial fractions be used for integrals, and if so, how?
4. How would you handle a case with complex roots in the denominator?
5. How can you verify the correctness of the partial fraction decomposition?

Tip: Partial fraction decomposition is often useful in integration, particularly for rational functions where direct integration isn't straightforward.