How was this calculated $(1-1/z)^{(\frac{-1}{3})}=(1+\frac{1}{3z}+\dots)$ *Background: a simplification step from an interesting longer problem here where the residue is needed so only the first two numbers are needed.
$$(1-\tfrac1z)^{-\frac13}=(1+\tfrac1{3z}+\cdots)$$
Was the likely step to compute the cube root: $(1+\frac{1}{z}+(\frac{1}{z})^2\dots)^{\frac{1}{3}}$?
If so, how do I do that?

Considering $$A=\frac{1}{\sqrt[3]{1-\frac{1}{z}}}$$ let for simplicity $z=\frac 1x$ to make $$A=(1-x)^{-1/3}$$ and use either the generalized binomial theorem or Taylor series around $x=0$. This will give $$A=1+\frac{x}{3}+\frac{2 x^2}{9}+\frac{14 x^3}{81}+\frac{35 x^4}{243}+\frac{91
   x^5}{729}+O\left(x^6\right)$$ Replace back $x$ by $\frac 1z$ to get $$A=1+\frac{1}{3 z}+\frac{2}{9 z^2}+\frac{14}{81 z^3}+\frac{35}{243 z^4}+\frac{91}{729
   z^5}+O\left(\frac{1}{z^6}\right)$$
For illustration purposes, using $z=2$, the above expansion gives $\frac{29317}{23328}\approx 1.25673$ while the exact value $\sqrt[3]{2}\approx 1.25992$. For sure, it will be better and better when $z$ will increase.
For example using $z=10$, the above expansion gives $\frac{75505741}{72900000}\approx 1.035744047$ while the exact value $\approx 1.035744169$.
