Double the sum, double the fun!

Notice that if we switch $a$ and $b$ , our sum will still converge to the same value. In other words, $\sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{6a^2}{b^6+a^4b^2} = \sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{6b^2}{a^6+b^4a^2}.$ If we let the value of this sum be $S$ , we have that $2S = \sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{6a^2}{b^6+a^4b^2} + \sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{6b^2}{a^6+b^4a^2}$ $= 6\sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{a^2}{b^2(a^4+b^4)} + \displaystyle \dfrac{b^2}{a^2(a^4+b^4)}$ $= 6\sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{a^4+b^4}{a^2b^2(a^4+b^4)}$ $= 6\sum_{a=1}^{\infty}\displaystyle\sum_{b=1}^{\infty} \dfrac{1}{a^2b^2} = 6\sum_{a=1}^{\infty}\dfrac{1}{a^2}\displaystyle\sum_{b=1}^{\infty} \dfrac{1}{b^2} = 6\bigg( \sum_{a=1}^{\infty}\dfrac{1}{a^2} \bigg)^2$ $= 6 \bigg( \dfrac{\pi^2}{6} \bigg)^2 \Longrightarrow 2S = \dfrac{\pi^4}{6}.$

Therefore, $S= \frac{\pi^4}{12}$ , and our final answer is

$12 + 4 = \boxed{16}$