Closed form for $\sum\limits_{n=2}^{\infty}\frac1{n^3-1}$

$$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b} $$ leads by partial fraction decomposition to $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)} = \frac{1}{(a-b)(b-c)(c-a)}\left[(b-c)\psi(a)+(c-a)\psi(b)+(a-b)\psi(c)\right] $$ In your case $a,b,c$ are the opposite of the roots of $(x+2)^3-1$, and I'll let you finish the computations: $\psi(1)=-\gamma$, but $\frac{1}{2}(5\pm i\sqrt{3})$ are not special arguments for $\psi(x)=\frac{d}{dx}\log\Gamma(x)$.

Partial fraction decomposition tells us that

$$\frac{1}{n^k-1}=\sum_{\omega^k=1} \frac{1}{n-\omega}\lim_{z\to\omega}\frac{z-\omega}{z^k-1}\overset{\text{L'H}}{=}\frac{1}{k}\sum_{\omega^k=1} \frac{\omega}{n-\omega}$$

and by the property that

$$\psi(z+1)=\psi(z)+\frac{1}{z}$$

we see that

$$\frac{1}{n^k-1}=\frac{1}{k}\sum_{\omega^k=1} \omega\Big(\psi(n-\omega+1)-\psi(n-\omega)\Big)$$

Then we may sum both sides from 2 to infinity:

$$\sum_{n=2}^{\infty}\frac{1}{n^k-1}=\frac{1}{k} \lim_{h \to \infty}\sum_{\omega^k=1} \omega\Big(\psi(h-\omega)-\psi(2-\omega)\Big)=-\frac{1}{k}\sum_{\omega^k=1} \omega \,\psi(2-\omega)$$

where the second equality is because $\psi(h)=\log(h)+O(1/h)$.

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Arriving at the WolframAlpha result:

Plug in $k=3$, and thus

$$\begin{align*}&\quad\sum_{n=2}^{\infty}\frac{1}{n^3-1}\\ &=-\frac{1}{3}\sum_{\omega^3=1}\omega\,\psi(2-\omega)\\ &=-\frac{1}{3}\sum_{\omega^3=1}\frac{\psi(2-\omega)}{\omega^2}\quad\big(\omega^3=1\big)\\ &=-\frac{1}{3}\sum_{(\omega+2)^3=1}\frac{\psi(-\omega)}{(\omega+2)^2}\quad\big(\omega\mapsto\omega+2\big)\\ &=-\frac{1}{3}\sum_{\omega^3+6\omega^2+12\omega+7=0}\frac{\psi(-\omega)}{\omega^2+4\omega+4} \end{align*}$$

which is what we wanted.