Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$

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

$$ S_1=\frac{1}{4}\cdot\frac{1}{3}\zeta(6)-\frac{1}{4}\cdot\frac{1}{24}\zeta(6)=\frac{7}{96}\zeta(6)=\frac{7}{96}\frac{\pi^6}{945}=\frac{\pi^6}{12960} $$