How do you write the partial fraction decomposition of the rational expression  (6x)/(x^3-8)?

Feb 29, 2016

$\frac{6 x}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)} = \frac{1}{x - 2} - \frac{x - 2}{{x}^{2} + 2 x + 4}$

Explanation:

Noting that from the difference of cubes formula,
${x}^{3} - 8 = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

We proceed to find the partial fraction decomposition.

$\frac{6 x}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)} = \frac{A}{x - 2} + \frac{B x + C}{{x}^{2} + 2 x + 4}$

$\implies 6 x = A \left({x}^{2} + 2 x + 4\right) + \left(B x + C\right) \left(x - 2\right)$

$= \left(A + B\right) {x}^{2} + \left(2 A - 2 B + C\right) x + \left(4 A - 2 C\right)$

Equating corresponding coefficients, we obtain the following system:

$\left\{\begin{matrix}A + B = 0 \\ 2 A - 2 B + C = 6 \\ 4 A - 2 C = 0\end{matrix}\right.$

Solve this using your favorite technique to find that it resolves to

$\left\{\begin{matrix}A = 1 \\ B = - 1 \\ C = 2\end{matrix}\right.$

Substituting these back in gives our final result.

$\frac{6 x}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)} = \frac{1}{x - 2} - \frac{x - 2}{{x}^{2} + 2 x + 4}$