Strange sum involving $\zeta (3)$

Question

$$\sum _{l=0}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^{k+l}}{2 (l+1)^3 \left((k+1)^3+(l+1)^3\right)}=\frac{9 \zeta (3)^2}{32}$$

I would like to prove the proposed equality, but I do not know where to start.

Answer 1

If $A = \sum_{n=1}^\infty\frac{(-1)^n}{n^3}$, then $A = -3\zeta(3)/4$. The required sum can be written as: $$ B = \sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^m(-1)^n}{n^3(m^3 + n^3)} = \sum_{m=1}^\infty\sum_{n=1}^\infty\frac{(-1)^m}{m^3}\left(\frac{(-1)^n}{n^3} - \frac{(-1)^{n}}{m^3 + n^3}\right) = A^2 - B, $$ whence the result.