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: 
The sum may be evaluated using the residue theorem. Basically, one considers the integral

$$\oint\_C dw \, \pi \, \cot{\pi w} \, f(w)$$

where $f(w) = (z^3-w^3)^{-1}$, and $C$ is a rectangular contour with vertices $$\pm (N+1/2) (1\pm i) $$

taken in a counterclockwise sense. To summarize: the contour integral is zero (the contributions vanish sufficiently fast as $N \to \infty$ - thanks, @markoriedel. See pages 76-78 of [this reference](http://www.maths.manchester.ac.uk/~cwalkden/complex-analysis/complex_analysis_part7.pdf) for a full derivation.)), but is also equal to the sum of the residues at the poles. The poles of the cotangent are at the integers, so that we have

$$\sum\_{n=-\infty}^{\infty} f(n) = - \pi \sum\_k \operatorname\*{Res}\_{w=w\_k} [\cot{\pi w} \, f(w)]$$

With the $f(w)$ we have, the poles are at $w=z$ and $w=z e^{\pm i 2 \pi/3}$. As all the poles are simple, we have

$$\operatorname\*{Res}\_{w=z} \frac{\pi \, \cot{\pi w}}{z^3-w^3} = -\frac{\cot{\pi z}}{3 z^2}$$

$$\operatorname\*{Res}\_{w=z e^{i 2 \pi/3}} \frac{\pi \, \cot{\pi w}}{z^3-w^3} = -\frac{\cot{\left (\pi z e^{i 2 \pi/3}\right )}}{3 z^2 e^{-i 2 \pi/3}}$$

$$\operatorname\*{Res}\_{w=z e^{-i 2 \pi/3}} \frac{\pi \, \cot{\pi w}}{z^3-w^3} = -\frac{\cot{\left (\pi z e^{-i 2 \pi/3}\right )}}{3 z^2 e^{i 2 \pi/3}}$$

So we just need to sum these up. The challenge is essentially evaluating the cotangent of the complex number, but it is straightforward and the rest is just algebra.

Note that we can express the result as

$$\frac{\pi}{3 z^2} \left [\cot{\pi z} + 2 \Re{\left [\cot{\left ( \pi z e^{i 2 \pi/3}\right ) e^{i 2 \pi/3}} \right ]} \right ] $$

which simplifies to

$$\frac{\pi}{3 z^2} \left [\cot (\pi z)+\frac{\sin (\pi z)+\sqrt{3} \sinh \left(\sqrt{3} \pi z\right)}{\cosh \left(\sqrt{3} \pi z\right)-\cos(\pi z)}\right ]$$

or, finally,

$$\sum\_{n=-\infty}^{\infty} \frac{1}{z^3-n^3} = \frac{2 \pi}{3 z^2} \frac{\sqrt{3} \sin (\pi z) \sinh \left(\sqrt{3} \pi z\right)+\cos(\pi z) \cosh \left(\sqrt{3} \pi z\right)-\cos (2 \pi z)}{2 \sin (\pi z) \cosh \left(\sqrt{3} \pi z\right)-\sin (2 \pi z)} $$

Here's a plot for $z \in [0.1,0.9]$ for both the expression above and a numerical approximation to the sum just to make sure:

![enter image description here](https://i.stack.imgur.com/mqLVr.jpg)
First rewrite it as:

$$\frac{1}{z^3} + \sum\_{n=1}^\infty \left(\frac{1}{z^3-n^3} + \frac{1}{z^3+n^3}\right) = \frac{1}{z^3} +\sum\_1^\infty \frac{2z^3}{z^6-n^6}$$

so we only need to find:

$$\sum\_{n=1}^\infty \frac{1}{z^6-n^6}$$

Now, if $\omega$ is a primitive cube root of unity, there are functions $a(u),b(u),c(u)$ such that for real $u\neq v$:

$$\frac{1}{u^3-v^3} = \frac{a(u)}{u-v} +\frac{b(u)}{\omega u -v} + \frac{c(u)}{\omega^2u-v}$$

Setting $u=z^2$ and $v=n^2$, and finding a solution $\zeta$ so that $\zeta^2=\omega$, then we have:

$$\frac{1}{z^6-n^6} = \frac{a(z^2)}{z^2 - n^2}+\frac{b\left((\zeta z)^2\right)}{(\zeta z)^2-n^2} + \frac{c\left((\zeta^2 z)^2\right)}{(\zeta^2 z)^2 -n^2}$$

So we can use your first result to find:

$$\sum\_{n=1}^\infty \frac{1}{z^6-n^6} = a(z^2)f(z) + b(\zeta^2z^2)f(\zeta z) + c(\zeta^4 z^2)f(\zeta^2 z)$$
where $$f(z)=\sum\_{n=1}^\infty \frac{1}{z^2-n^2} = \frac{\pi z\cot(\pi z)-1}{2z^2}$$
That seems a bit messy, and requires jumping into complex numbers...