complex analysis Find the residue of the function $f(z) = z/ (8-z^3)$ at $z = 2.$
I tried with the formula $\lim_{z\rightarrow a} \frac{1}{(m-1)!} \frac{ d^{m-1}}{ dz^{m-1}} (z-a)^m f(z)$. but it becomes very tedious. does it have any other formula for calculating these types of functions?
 A: One may just start with
$$
f(z):=\frac{z}{8-z^3}=\frac{z}{2^3-z^3}=\frac{z}{(2-z)(4+2z+z^2)}
$$ giving, as $z \to 2$,

$$
(z-2)f(z)=-\frac{z}{4+2z+z^2} \to \color{red}{-\frac16}.
$$

A: Let $g(z)=8-z^3$ and $z_0=2$. Then it holds that $g(z_0)=0$ and $g'(z_0)\neq 0$ as well as $z\mapsto z$ being holomorphic. Thus we can apply
$$\operatorname{Res}_{z=z_0}\left(\frac{f(z)}{g(z)}\right)=\frac{f(z_0)}{g'(z_0)}.$$
For your problem you get
$$\operatorname{Res}_{z=2}\left(\frac{z}{8-z^3}\right)=\frac{2}{-3\cdot 2^2}=-\frac{1}{6}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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Lets $\ds{\mathcal{R} \equiv \braces{w\ \mid\ w^{3} = 8}}$:
\begin{align}
{z \over 8 - z^{3}} & =
-z\pars{{1 \over 24}\sum_{r\ \in\ \mathcal{R}}{r \over z - r}} =
-\,{1 \over 24}\sum_{r\ \in\ \mathcal{R}}{z\,r \over z - r}
\end{align}
Then, $\ds{-\,{1 \over 24}\,2\times 2 = \color{#f00}{-\,{1 \over 6}}}$.
