An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$?

We have $\begin{align} \displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2} & = \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{(n+k)^3 - n^3-k^3}{n^3k^3(n+k)^3} \\ &= \frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{n^3(n+k)^3} - \frac{1}{3}\sum\limits_{n,k=1}^{\infty}\frac{1}{k^3(n+k)^3} \\ &=\frac{1}{3}\sum\limits_{n,k=1}^{\infty} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n>k} \frac{1}{n^3k^3} - \frac{1}{3}\sum\limits_{n<k} \frac{1}{n^3k^3} \\&= \frac{1}{3}\sum\limits_{n=k} \frac{1}{n^3k^3} = \frac{1}{3}\zeta(6) \end{align}$

We have, $\displaystyle B_{2n}(x) = (-1)^{n-1}\frac{2(2n)!}{(2\pi)^{2n}} \sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}$

Cubing and integrating both sides,

\begin{align*}&\int_0^1 \left((-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}(x)\right)^3\,dx \\&= \int_0^1 \left(\sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}\right)^3\,dx \\&= \sum\limits_{k_1,k_2,k_3 = 1}^{\infty} \int_0^1 \frac{\cos \left(2\pi k_1x \right)\cos \left(2\pi k_2x \right)\cos \left(2\pi k_3x \right)}{k_1^{2n}k_2^{2n}k_3^{2n}}\,dx\\&= \frac{1}{4}\sum\limits_{k_1,k_2,k_3 = 1}^{\infty} \dfrac{\displaystyle \sum\limits_{\{\pm\}} \int_0^1 \cos \left(2\pi (k_1\pm k_2 \pm k_3)x \right)\,dx}{k_1^{2n}k_2^{2n}k_3^{2n}}\\&= \frac{3}{4}\sum\limits_{k_1,k_2= 1}^{\infty} \frac{1}{k_1^{2n}k_2^{2n}(k_1+k_2)^{2n}}\end{align*}

Since, $\displaystyle \int_0^1 \cos \left(2\pi (k_1 + k_2 - k_3)x \right)\,dx = \begin{cases}1 & \text{ when } k_3 = k_1 + k_2\\ 0 &\text{ otherwise }\end{cases}$

The case $n = 1$,

$$\sum\limits_{k_1,k_2= 1}^{\infty} \frac{1}{k_1^{2}k_2^{2}(k_1+k_2)^{2}} = \frac{4\pi^6}{3}\int_0^1 \left(B_2(x)\right)^3\,dx = \frac{1}{3}\frac{\pi^6}{945} = \frac{1}{3}\zeta(6)$$