Question

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

Answer 1

`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`