Evaluating $\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{k-1}}{n^2\left(k^2-2n^2 \right)}$

Question

I am trying to evaluate the following series: $$\sum_{n=1}^\infty\frac{1}{n^3\sin\left(\sqrt{2}\pi n \right)}\tag{a} $$ where, using the well-known result: $$\frac{1}{\sin\left(\pi x\right)}=\frac{2x}{\pi}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^2-x^2}+\frac{1}{\pi x}\tag{1} $$ connecting $(1)$ in $(a)$ and $x\to \sqrt{2}n$ got the step: $$\sum_{n=1}^\infty\frac{1}{n^3\sin\left(\sqrt{2}\pi n \right)}=\frac{2\sqrt{2}}{\pi}\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{k-1}}{n^2\left(k^2-2n^2 \right)}+\frac{1}{\sqrt{2}\pi}\zeta(4)\tag{b} $$ I tried to apply partial fractions in the double series above, then i found that: $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{k-1}}{n^2\left(k^2-2n^2 \right)}=\frac54\zeta(4)+2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{k-1}}{k^2\left(k^2-2n^2\right)}\tag{2} $$ To conclude, I do not know to what extent this last step can help to solve the Double Series. Maybe I'm not seeing the obvious. It will be interesting to see some approach to resolve it. At Wolfram, she is: $$\therefore\ \sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(-1)^{k-1}}{n^2\left(k^2-2n^2 \right)}=-\frac{17}{16}\zeta(4). $$