# How do you express 1/[(x^3)-1] in partial fractions?

Mar 9, 2016

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

#### Explanation:

Sticking with Real coefficients, we can factor the denominator as:

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

Hence we are looking for a decomposition of the form:

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

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

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

So equating coefficients we get three simultaneous equations:

$A + B = 0$

$A - B + C = 0$

$A - C = 1$

Adding all of these three equations together we find:

$3 A = 1$

Hence $A = \frac{1}{3}$, $B = - \frac{1}{3}$ and $C = - \frac{2}{3}$

So:

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