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

Jul 27, 2016

There is no breakdown into simpler fractions with Real coefficients, but with Complex coefficients we find:

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

#### Explanation:

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

The zeros of ${x}^{2} + x + 1$ are the non-Real Complex cube roots of $1$, $\omega$ and ${\omega}^{2} = \overline{\omega}$

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

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

$= \frac{A}{x - {\omega}^{2}} + \frac{B}{x - \omega}$

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

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

Equating coefficients:

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

Substituting $B = - A$ in the second of these equations we get:

$A \omega - A {\omega}^{2} = - 1$

Rearranging, we find:

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

So:

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

$= \frac{\sqrt{3} i}{3 \left(x - {\omega}^{2}\right)} - \frac{\sqrt{3} i}{3 \left(x - \omega\right)}$

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