TITLE: How can I prove $\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$
QUESTION [10 upvotes]: Can the residue theorem prove this?
$$\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$$

REPLY [6 votes]: Even though I agree that the solution by @math110 is the best one, I want to add that this can be done using the theory of residues, using the common trick with the cotangent function. I leave it to you to fill in the details, if necessary.
We let
$$
f(z)=\frac{\pi\cot \pi z}{z^3(z+1)^3}.
$$
This function $f$ has poles at all integers. For integers $n\not\in\{-1,0\}$, the residue of $f$ at $n$ is $1/(n(n+1))^3$. For $n\in\{-1,0\}$, the residue of $f$ at $n$ is $\pi^2-10$.
Integrating over the (oriented) curve consisting of the boundary of 
$$
C_N=\{z\in\mathbb{C}~|~-N-3/2<\text{Re}\, z<N+1/2,\ -N-3/2<\text{Im}\, z<N+1/2\},
$$
you will find that (note the symmetry in the line $\text{Re}\,z=-1/2$!), as $N\to+\infty$,
$$
0=(\pi^2-10)+(\pi^2-10)+2\sum_{n=1}^{+\infty}\frac{1}{n^3(n+1)^3}.
$$