I can't figure it out how to end this problem:
Find the Laurent Series of the function $$f(z)=\frac{1}{(z^3+1)(z^3-2)}$$ valid in $$A=\{z \in \mathbb{C} : 1 < |z|<2^{\frac{1}{3}}$$
I know that:
$$ f(z)=\frac{1}{(z^3+1)(z^3-2)}=\frac{1}{3}*\frac{1}{(z^3+1)}-\frac{1}{3}*\frac{1}{(z^3-2)} $$
$$ \frac{1}{z^3}*\frac{1}{1+1/z^3}= \frac{1}{(z^3+1)} $$
Best Answer
If $|z|>1$, then\begin{align}\frac1{z^3+1}&=\frac1{1-(-z^3)}\\&=-\sum_{n-\infty}^{-1}(-z^3)^n\\&=\sum_{n=-\infty}^{-1}(-1)^{n+1}z^{3n},\end{align}and if $|z|<\sqrt[3]2$, then\begin{align}\frac1{z^3-2}&=-\frac12\frac1{1-z^3/2}\\&=-\frac12\sum_{n=0}^\infty\left(\frac{z^3}2\right)^n\\&=-\sum_{n=0}^\infty\frac{z^{3n}}{2^{n+1}}.\end{align}