Inverse Sum Cubed

Note first that $\displaystyle\sum_{k=0}^{n} k^{3} = \left(\dfrac{n(n + 1)}{2}\right)^{2}.$ So the given sum can be expressed as

$4\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{n(n + 1)}\right)^{2} = 4\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)^{2} = 4\sum_{n=1}^{\infty} \left(\dfrac{1}{n^{2}} + \dfrac{1}{(n + 1)^{2}} - \dfrac{2}{n(n + 1)}\right) =$

$4\displaystyle\left(2\sum_{n=1}^{\infty} \left(\dfrac{1}{n^{2}}\right) - 1\right) - 8\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right) = 4*\left(\dfrac{2\pi^{2}}{6} - 1\right) - 8*1 = \dfrac{4\pi^{2}}{3} - 12.$

Thus $a + b + c = 4 + 3 + 12 = \boxed{19}.$

Note that $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{2}} = \dfrac{\pi^{2}}{6}$ is the solution to the Basel problem , and $\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)$ is a telescoping sum where all the terms cancel pairwise except the very first, namely $\dfrac{1}{1} = 1.$