Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$

Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\displaystyle \sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\displaystyle \sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$
but nothing seems to lead to anything!

HINT:
Use the partial fraction decomposition
$$\frac{1}{n^3(n+1)^3} = \left(\frac{1}{n^3} - \frac{1}{(n + 1)^3}+ \frac{6}{n} - \frac{6}{n+1}\right) - \left(\frac{3}{n^2} +  \frac{3}{(n + 1)^2}\right) $$
The sum equals $(1+6) -( 3 \zeta(2) +3 \zeta(2)- 3)= 10 - \pi^2$.
