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

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Answer 1 · 2017-01-07T10:58:27

The answer is $=(1/9)/x+(1/3)/(x-3)^2+(-1/9)/(x-3)$

Let's factorise the denominator

$x^3-6x^2+9x=x(x^2-6x+9)= x(x-3)^2$

Therefore,

$1/(x^3-6x^2+9x)=1/(x(x-3)^2)$

$=A/x+B/(x-3)^2+C/(x-3)$

$=(A(x-3)^2+Bx+Cx(x-3))/(x(x-3)^2)$

So,

$1=A(x-3)^2+Bx+Cx(x-3)$

Let $x=0$ , $=>$ , $1=9A$

Coefficients of $x^2$ ,

$0=A+C$ , $=>$ , $C=-A=-1/9$

Let $x=3$ , $=>$ , $1=3B$

Therefore,

$1/(x^3-6x^2+9x)=(1/9)/x+(1/3)/(x-3)^2+(-1/9)/(x-3)$

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