We have the partial fraction decomposition 
$$\frac{ca^2}{k^3+a^3}=\frac{\omega}{k+a/\omega }+\frac{-\omega -1}{k+a}+\frac{1}{k+a \omega},$$
where $c:=-(\omega +1)$ and $\omega:=e^{i\pi/3}$. 
Also, 
$$\sum_{k=1}^n\frac1{k+b}=\ln n-\psi(1+b)+o(1)$$
(as $n\to\infty$), where $\psi$ is the digamma function. 
Collecting the pieces, we get 
$$\sum_{k=1}^\infty\frac1{k^3+a^3}
=\frac1{ca^2}\,
\left((\omega +1) \psi(1+a)-\omega  \psi\left(1+a/\omega\right)-\psi(1+a \omega)\right).$$
(For $a\to0$, this is $-\psi ^{(2)}(1)/2+O(a)=\zeta(3)+O(a)$.)