# How do you express  1 / (x^6 - x^3) in partial fractions?

Aug 30, 2016

$\frac{1}{{x}^{6} - {x}^{3}} = \frac{1}{3 \left(x - 1\right)} - \frac{x + 2}{3 \left({x}^{2} + x + 1\right)} - \frac{1}{x} ^ 3$

#### Explanation:

First note that:

$\frac{1}{{t}^{2} - t} = \frac{1}{t \left(t - 1\right)} = \frac{t - \left(t - 1\right)}{t \left(t - 1\right)} = \frac{1}{t - 1} - \frac{1}{t}$

So putting $t = {x}^{3}$ we find:

$\frac{1}{{x}^{6} - {x}^{3}} = \frac{1}{{x}^{3} - 1} - \frac{1}{x} ^ 3$

We cannot simplify $\frac{1}{x} ^ 3$, but we can work on the other term...

$\frac{1}{{x}^{3} - 1}$

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

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

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

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

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

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

So:

$\frac{1}{{x}^{6} - {x}^{3}} = \frac{1}{3 \left(x - 1\right)} - \frac{x + 2}{3 \left({x}^{2} + x + 1\right)} - \frac{1}{x} ^ 3$