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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 $$\sum_{k=1}^n\frac1{k^3+a^3} =\frac1{ca^2}\, \left((\omega +1) \psi(1+a)-\omega \psi\left(1+a/\omega\right)-\psi(1+a \omega)\right).$$