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

1 Answer
Aug 30, 2016

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

Explanation:

First note that:

#1/(t^2-t) = 1/(t(t-1)) = (t - (t-1))/(t(t-1)) = 1/(t-1)-1/t#

So putting #t = x^3# we find:

#1/(x^6-x^3) = 1/(x^3-1)-1/x^3#

We cannot simplify #1/x^3#, but we can work on the other term...

#1/(x^3-1)#

#= 1/((x-1)(x^2+x+1))#

#= 1/3*3/((x-1)(x^2+x+1))#

#= 1/3*((x^2+x+1) - (x^2+x-2))/((x-1)(x^2+x+1))#

#= 1/3*((x^2+x+1) - (x-1)(x+2))/((x-1)(x^2+x+1))#

#= 1/3*(1/(x-1)-(x+2)/(x^2+x+1))#

#= 1/(3(x-1))-(x+2)/(3(x^2+x+1))#

So:

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