# How do you express  2 / (x^3 + 1) in partial fractions?

Jun 18, 2016

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

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

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.

#### Explanation:

Let $\omega$ denote the primitive Complex cube root of $1$.

That is:

$\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$

Then:

${\omega}^{2} = \overline{\omega} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

${\omega}^{3} = 1$

$1 + \omega + {\omega}^{2} = 0$

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

We find:

$\frac{1}{x + 1} + \frac{\omega}{x + \omega} + {\omega}^{2} / \left(x + {\omega}^{2}\right)$

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

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

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

Therefore:

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

If we want a partial fraction decomposition with only Real coefficients, then combine the last two terms so:

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

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