# How do you express x^4/(x-1)^3 in partial fractions?

##### 1 Answer
Nov 3, 2016

${x}^{4} / {\left(x - 1\right)}^{3} = x + 3 + \frac{6}{x - 1} + \frac{4}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3$

#### Explanation:

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

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = \frac{\left({x}^{4} - 3 {x}^{3} + 3 {x}^{2} - x\right) + \left(3 {x}^{3} - 3 {x}^{2} + x\right)}{{x}^{3} - 3 {x}^{2} + 3 x - 1}$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + \frac{3 {x}^{3} - 3 {x}^{2} + x}{{x}^{3} - 3 {x}^{2} + 3 x - 1}$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + \frac{\left(3 {x}^{3} - 9 {x}^{2} + 9 x - 3\right) + \left(6 {x}^{2} - 8 x + 3\right)}{{x}^{3} - 3 {x}^{2} + 3 x - 1}$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + 3 + \frac{6 {x}^{2} - 8 x + 3}{{x}^{3} - 3 {x}^{2} + 3 x - 1}$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + 3 + \frac{A}{x - 1} + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1} ^ 3$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + 3 + \frac{A {\left(x - 1\right)}^{2} + B \left(x - 1\right) + C}{x - 1} ^ 3$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + 3 + \frac{A \left({x}^{2} - 2 x + 1\right) + B \left(x - 1\right) + C}{x - 1} ^ 3$

$\textcolor{w h i t e}{{x}^{4} / {\left(x - 1\right)}^{3}} = x + 3 + \frac{A {x}^{2} + \left(- 2 A + B\right) x + \left(A - B + C\right)}{x - 1} ^ 3$

Equating coefficients, we find:

$\left\{\begin{matrix}A = 6 \\ - 2 A + B = - 8 \\ A - B + C = 3\end{matrix}\right.$

Hence:

$\left\{\begin{matrix}A = 6 \\ B = 4 \\ C = 1\end{matrix}\right.$

So:

${x}^{4} / {\left(x - 1\right)}^{3} = x + 3 + \frac{6}{x - 1} + \frac{4}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3$