How can I prove $\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$ Can the residue theorem prove this?
$$\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$$
 A: 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 $N\to+\infty$,
$$
0=(\pi^2-10)+(\pi^2-10)+2\sum_{n=1}^{+\infty}\frac{1}{n^3(n+1)^3}.
$$
A: Hint
$$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$
and $$(a-b)^3=(a^3-b^3)-3ab(a-b)$$
so
\begin{align*}\dfrac{1}{n^3(n+1)^3}&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-3\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\left(\dfrac{1}{n}-\dfrac{1}{(n+1)}\right)\\
&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-\dfrac{3}{n^2}+6\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-\dfrac{3}{(n+1)^2}
\end{align*}
and use well kown
$$\zeta{(2)}=\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$ can solve it
