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 $c:=3(\omega-1)$ 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, for $a\in(-1,\infty)\setminus\{0\}$ we get 
$$s(a):=\sum_{k=1}^\infty\frac1{k^3+a^3}
=\frac1{ca^2}\,
\left((1-\omega) \psi(1+a)+\omega\psi\left(1-a/\omega\right)
-\psi(1-a \omega)\right).$$
(For $a\to0$, $s(a)=-\psi ^{(2)}(1)/2+O(a)=\zeta(3)+O(a)$.)

Here is the graph $\{(a,s(a))\colon0<a\le1\}$, with $s(0)=\zeta(3)=1.2020\ldots$: 

[![enter image description here][1]][1]

(I am not geting instability.)

  [1]: https://i.sstatic.net/kkbB7.png