Question: I want to express $$\sum\_{n=-\infty}^\infty\dfrac{1}{z^3-n^3}$$ in closed form.

I know that $$\pi z\cot(\pi z)=1+2z^2\sum\_{n=1}^\infty\dfrac{1}{z^2-n^2}$$ which looks close, but I don’t know how to use it.

Answer: 
Assuming you are referring to the book by Ahlfors, here is a method relying on it:

Starting with a partial fraction decomposition w.r.t. $n$ we find

$$\frac{1}{z^3-n^3}=\frac{ \rho\_3^0}{3z^2} \frac{1}{ \rho\_3^0 z-n}+\frac{\rho\_3^1}{3z^2} \frac{1}{ \rho\_3 z-n}+\frac{\rho\_3^2}{3z^2} \frac{1}{ \rho\_3^2 z-n} $$ where $\rho\_3=\exp(2 \pi i/3)$. Using the formula $$ \pi \cot( \pi z)= \lim\_{m \to \infty} \sum\_{n=-m}^m \frac{1}{z-n} $$
from page 189 three times should give you the desired result. I will add that since the series converges in the usual sense, it also converges w.r.t. this "principal value summation".
Here is a closed form by Maple

> 
> $$ \frac{2}{3}\,{\frac {\pi \, \left( -\cos \left( 2\,\pi \,z \right) +\sqrt {3}
> \sin \left( \pi \,z \right) \sinh \left( \pi \,z\sqrt {3} \right) +
> \cos \left( \pi \,z \right) \cosh \left( \pi \,z\sqrt {3} \right) 
>  \right) }{{z}^{2} \left( 2\,\sin \left( \pi \,z \right) \cosh \left( 
> \pi \,z\sqrt {3} \right) -\sin \left( 2\,\pi \,z \right) \right) }}.$$
> 
> 
>