Don't forget 2015

$\large \displaystyle \sum_{k=2}^{63} \dfrac{k^3-k}{k^6-1}=\dfrac{k(\color{#D61F06}{k^2-1})}{(\color{#D61F06}{k^2-1})(k^4+k^2+1)}$ $=\dfrac{1}{2} \displaystyle \sum_{k=2}^{63} \dfrac{1}{k^2-k+1}- \dfrac{1}{k^2+k+1}~~\color{#20A900}{(Telescopic~Series)}$ $=\dfrac{1}{2}(\dfrac{1}{3}-\dfrac{1}{4033})$ $=\dfrac{\color{goldenrod}{2015}}{12099}$ $\Large 2015+12099=\boxed{\color{darkviolet}{\boxed{14114}}}$

$\dfrac{k^3-k}{k^6-1}=\dfrac{k(k^2-1)}{(k^3)^2-1} =\dfrac{k(k-1)(k+1)}{(k^3+1)(k^3-1)} =\dfrac{k(k-1)(k+1)}{(k+1)(k^2+1-k)(k-1)(k^2+1+k)} \\ =\dfrac{k}{(k^2+1-k)(k^2+1+k)} \\ =\dfrac{1}{2}\left(\dfrac{1}{k^2+1-k}-\dfrac{1}{k^2+1+k}\right)$ $\begin{aligned} \displaystyle \sum_{k=2}^{63} \dfrac{k^3-k}{k^6-1} & =\dfrac{1}{2}\displaystyle \sum_{n=2}^{63} \dfrac{1}{k^2+1-k}-\dfrac{1}{k^2+1+k} \\ & =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\ldots- \dfrac{1}{63^2+1+63}\right) \\ & =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{4033}\right) =\dfrac{1}{2}\cdot\dfrac{4030}{12099} \\ &=\boxed{\dfrac{2015}{12099}} \end{aligned}$