We have the partial fraction decomposition 
$$\frac c{k^3+a^3}=\frac{\omega}{k+a/\omega }+\frac{-\omega -1}{k+a}+\frac{1}{k+a \omega},$$
where $c:=a^2 (\omega -1)^2 (\omega +1)/\omega\,$ and $\omega:=e^{i\pi/3}$. 
Also, 
$$\sum_{k=1}^n\frac1{k+b}=\ln n-\psi(b+1)+o(1)$$
(as $n\to\infty$), where $\psi$ is the digamma function. 
Collecting the pieces, we get 
$$\sum_{k=1}^n\frac1{k^3+a^3}
=\frac{\omega ^{5/4} }{a^2\sqrt{3} }\,
\left((\omega +1) \psi(a+1)-\omega  \psi\left(1-a \omega ^2\right)-\psi(a \omega +1)\right).$$