find coefficient of $x^{50}$ Let $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}$then find the coefficient of term $x^{50}$ in $(f(x))^3$.I think that we can set $$(f(x))^3=\frac{a}{(1+x)^3}+\frac{b}{(1+x^2)^3}+\frac{c}{(1+x^4)^3}$$ and find a,b and c then use Taylor seri .

We have $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}=\frac{1-x}{1-x^8}=(1-x)(1-x^8)^{-1}$ if $|x^8|<1$.
Now note that when $|x|<1$ then \begin{align*}
(1-x)^{-1}=&\sum\limits_{n\geq 0}x^n\\
-(1-x)^{-2}=&\sum\limits_{n\geq 1}nx^{n-1}\\
+2(1-x)^{-3}=&\sum\limits_{n\geq 2}n(n-1)x^{n-2}=\sum\limits_{n\geq 0}(n+2)(n+1)x^{n}\\
(1-x)^{-3}=&\sum\limits_{n\geq 0}\frac{(n+2)(n+1)}{2}x^{n}
\end{align*}
which means $f(x)^3=(1-x)^3(1-x^8)^{-3}=(1-3x+3x^2-x^3)\sum\limits_{n\geq 0}\frac{(n+2)(n+1)}{2}x^{8n}$
On the RHS, we have the terms of the forms $x^{8n}, x^{8n+1}, x^{8n+2}, x^{8n+3}$. Since none of $8n, 8n+1, 8n+3$ but $8n+2$ provide 50 (for $n=6$) so the required coefficient of $x^{50}$ is $3\frac{8.7}{2}=84$
