# How do you write the partial fraction decomposition of the rational expression  (x^3 - 2) / (x^4 - 1)?

Dec 5, 2015

$\frac{{x}^{3} - 2}{{x}^{4} - 1} = \frac{x + 2}{2 \left({x}^{2} + 1\right)} + \frac{3}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}$

#### Explanation:

First, note that we can factor the denominator as

${x}^{4} - 1 = \left({x}^{2} + 1\right) \left({x}^{2} - 1\right) = \left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 2\right)$

Then, using partial fraction decomposition,

$\frac{{x}^{3} - 2}{{x}^{4} - 1} = \frac{A x + B}{{x}^{2} + 1} + \frac{C}{x + 1} + \frac{D}{x - 1}$

Multiplying through by $\left({x}^{2} + 1\right) \left(x + 1\right) \left(x - 1\right)$ gives

${x}^{3} - 2$

$= \left(A x + B\right) \left({x}^{2} - 1\right) + C \left({x}^{2} + 1\right) \left(x - 1\right) + D \left({x}^{2} + 1\right) \left(x + 1\right)$

$= \left(A + C + D\right) {x}^{3} + \left(B - C + D\right) {x}^{2} + \left(- A + C + D\right) x + \left(- B - C + D\right)$

Equating coefficients then gives us the system

$\left\{\begin{matrix}A + C + D = 1 \\ B - C + D = 0 \\ - A + C + D = 0 \\ - B - C + D = - 2\end{matrix}\right.$

Solving this gives us
$\left\{\begin{matrix}A = \frac{1}{2} \\ B = 1 \\ C = \frac{3}{4} \\ D = - \frac{1}{4}\end{matrix}\right.$

Substituting back, we get the final result.

$\frac{{x}^{3} - 2}{{x}^{4} - 1} = \frac{\frac{1}{2} x + 1}{{x}^{2} + 1} + \frac{\frac{3}{4}}{x + 1} + \frac{- \frac{1}{4}}{x - 1}$

$= \frac{x + 2}{2 \left({x}^{2} + 1\right)} + \frac{3}{4 \left(x + 1\right)} - \frac{1}{4 \left(x - 1\right)}$