An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? While Solving some integral problem, I encountered the following infinite series:
$$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$
I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.

We have $\begin{align} \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2} & = \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{(n+k)^3 - n^3-k^3}{n^3k^3(n+k)^3} \\ &= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{n^3(n+k)^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{k^3(n+k)^3} \tag{*}\\ &=\frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n=1}^\infty\sum\limits_{k = n+1}^\infty \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{k=1}^\infty\sum\limits_{n=k+1}^\infty \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{1 \le n < k <\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{1 \le k < n < \infty} \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n=k=1}^\infty \frac{1}{n^3k^3} =  \frac{1}{3}\zeta(6) \end{align}$
where, in line $(*)$ we reindexed the summation with the change of variables $n+k \mapsto k$ in the second sum and $n+k \mapsto n$ in the third sum.
