# How do you write the partial fraction decomposition of the rational expression (x^2 − x + 1 )/( x^3 − x^2 + x − 1)?

Aug 7, 2016

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

#### Explanation:

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

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

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

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

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

Equating coefficients, we get:

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

Hence:

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

So:

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