# Partial fractions

I'm trying to do a question that requires the expansion of the following using partial fractions:
$$f(z)=\frac{1}{(1+z^3)^2}$$.
The fact that the bottom is squared is throwing me off for some reason... I've factorized the bottom, but I'm not sure whether I should use the complex roots or not, or even if it's possible without using complex roots.
Any help would be appreciated.

Hey there,
The trick is to use identity 1+Z^3=(1+z)(1-Z+Z^2) and then just square this expression and do the partial fraction as usual.
Hope this helps)

HallsofIvy
$$\frac{1}{(1+z^3)^2}= \frac{1}{(z+1)^2(z^2-z+1)^2}$$
$$\frac{A}{z+1}+ \frac{B}{(z+1)^2}+ \frac{Cx+D}{z^2- z+1}+ \frac{Ex+F}{(z^2- z+1)^2}$$
There was an earlier question about the "principal part" or "Laurent series" for $1/(z^2+1)^2$ which I answered by reducing to partial fractions with complex coefficients. Is this related to that thread?