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)#