<s>[INST] Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$ As a follow up of this nice question I am interested in 
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m n\left(m^2-n^2\right)^2}
$$
Furthermore, I would be also very grateful for a solution to
$$
S_2=\sum_{m=1}^{\infty}\sum_{n=m+1}^{\infty}\frac{  1}{m n\left(m^2-n^2\right)^2}
$$
Following my answer in the question mentioned above and the numerical experiments of @Vladimir Reshetnikov it's very likely that at least 
$$
S_1+S_2 = \frac{a}{b}\pi^6
$$
I think both sums may be evaluated by using partial fraction decomposition and the integral representation of the Polygamma function but I don't know how exactly and I guess there could be a much more efficient route.
 [/INST] Clearly, $S_1$=$S_2$ (this can be shown by reversing the order of summation, as was noted above).
Using
$$
\frac{  1}{m n\left(m^2-n^2\right)^2}=\frac{  (m+n)^2-(m-n)^2}{4 m^2 n^2\left(m^2-n^2\right)^2}
$$
we get
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m n\left(m^2-n^2\right)^2}=\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m^2 n^2\left(m+n\right)^2},
$$
and after reversing the order of summation in the first sum
$$
S_1=\frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{  1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m^2 n^2\left(m+n\right)^2}=\\
\frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{  1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m^2 n^2\left(m+n\right)^2}. \qquad\qquad (1)
$$
Let's introduce a third sum
$$
S_3=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{  1}{m^2 n^2\left(m+n\right)^2}
=\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{  1}{m^2 n^2\left(m+n\right)^2}=\\
\frac{1}{2}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{  1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{8}\zeta(6).
$$
Using An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? we get 
$$
S_3=\frac{1}{2}\cdot\frac{1}{3}\zeta(6)-\frac{1}{8}\zeta(6)=\frac{1}{24}\zeta(6).\qquad\qquad\qquad (2)
$$
From (1) and (2) we get
$$
S_1=\frac{1}{4}\cdot\frac{1}{3}\zeta(6)-\frac{1}{4}\cdot\frac{1}{24}\zeta(6)=\frac{7}{96}\zeta(6)=\frac{7}{96}\frac{\pi^6}{945}=\frac{\pi^6}{12960}
$$
</s>