Question

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

Answer 1

`2/(x^3+1) = 2/(3(x+1))+(2omega)/(3(x+omega))+(2omega^2)/(3(x+omega^2))`

`=2/(3(x+1))+(-2x+4)/(3(x^2-x+1))`

where `omega=-1/2+sqrt(3)/2i` is the primitive Complex cube root of `1`.

Explanation:

Let `omega` denote the primitive Complex cube root of `1`.

That is:

`omega = -1/2+sqrt(3)/2i`

Then:

`omega^2 = bar(omega) = -1/2-sqrt(3)/2i`

`omega^3 = 1`

`1 + omega + omega^2 = 0`

`x^3+1 = (x+1)(x+omega)(x+omega^2)`

We find:

`1/(x+1)+omega/(x+omega)+omega^2/(x+omega^2)`

`=((x+omega)(x+omega^2)+omega(x+1)(x+omega^2)+omega^2(x+1)(x+omega))/(x^3+1)`

`=((1+omega+omega^2)x^2+2(1+omega+omega^2)x+3)/(x^3+1)`

`=3/(x^2+1)`

Therefore:

`2/(x^3+1) = 2/(3(x+1))+(2omega)/(3(x+omega))+(2omega^2)/(3(x+omega^2))`

If we want a partial fraction decomposition with only Real coefficients, then combine the last two terms so:

`=2/(3(x+1))+(2omega(x+omega^2)+2omega^2(x+omega))/(3(x+omega)(x+omega^2))`

`=2/(3(x+1))+(-2x+4)/(3(x^2-x+1))`