# How do you express (8x-1)/(x^3 -1) in partial fractions?

Dec 8, 2016

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{7}{3 \left(x - 1\right)} - \frac{7 x - 10}{3 \left({x}^{2} + x + 1\right)}$

#### Explanation:

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

So:

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + x + 1}$

$\textcolor{w h i t e}{\frac{8 x - 1}{{x}^{3} - 1}} = \frac{A \left({x}^{2} + x + 1\right) + \left(B x + C\right) \left(x - 1\right)}{{x}^{3} - 1}$

$\textcolor{w h i t e}{\frac{8 x - 1}{{x}^{3} - 1}} = \frac{\left(A + B\right) {x}^{2} + \left(A - B + C\right) x + \left(A - C\right)}{{x}^{3} - 1}$

Equating coefficients we find:

$\left\{\begin{matrix}A + B = 0 \\ A - B + C = 8 \\ A - C = - 1\end{matrix}\right.$

Adding all three equations, we find:

$3 A = 7$

So:

$A = \frac{7}{3}$

From the first equation we can deduce:

$B = - A = - \frac{7}{3}$

From the third equation:

$C = A + 1 = \frac{7}{3} + 1 = \frac{10}{3}$

So:

$\frac{8 x - 1}{{x}^{3} - 1} = \frac{7}{3 \left(x - 1\right)} - \frac{7 x - 10}{3 \left({x}^{2} + x + 1\right)}$