# How do you express 1/(x^3-6x^2+9x) in partial fractions?

Jan 7, 2017

The answer is $= \frac{\frac{1}{9}}{x} + \frac{\frac{1}{3}}{x - 3} ^ 2 + \frac{- \frac{1}{9}}{x - 3}$

#### Explanation:

Let's factorise the denominator

${x}^{3} - 6 {x}^{2} + 9 x = x \left({x}^{2} - 6 x + 9\right) = x {\left(x - 3\right)}^{2}$

Therefore,

$\frac{1}{{x}^{3} - 6 {x}^{2} + 9 x} = \frac{1}{x {\left(x - 3\right)}^{2}}$

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

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

So,

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

Let $x = 0$, $\implies$, $1 = 9 A$

Coefficients of ${x}^{2}$,

$0 = A + C$, $\implies$, $C = - A = - \frac{1}{9}$

Let $x = 3$, $\implies$, $1 = 3 B$

Therefore,

$\frac{1}{{x}^{3} - 6 {x}^{2} + 9 x} = \frac{\frac{1}{9}}{x} + \frac{\frac{1}{3}}{x - 3} ^ 2 + \frac{- \frac{1}{9}}{x - 3}$