If $x$ satisfies $x^{37}=1,x\ne 1$, I want to evaluate
$$\sum_{n=1}^{36}\frac{n}{(1+x^n+x^{2n}+x^{3n})^3} = \sum_{n=1}^{36}n\left(\frac{x^n-1}{x^{4n}-1}\right)^3$$
Since $x^{37}=1,$
$$\sum_{n=1}^{36}n\left(\frac{x^n-1}{x^{4n}-1}\right)^3 $$ $$=\sum_{n=1}^{36}n\left(\frac{(x^{37})^{3n}x^n-1}{x^{4n}-1}\right)^3 $$ $$=\sum_{n=1}^{36}n\left(\frac{x^{112n}-1}{x^{4n}-1}\right)^3 $$ $$=\sum_{n=1}^{36}n(x^{108n}+x^{104n}+...+x^{4n}+1)^3 $$
But I have no idea how to continue.