<s>[INST] if $x^5=1$ with $x\neq 1$ then find value of $\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$ 
if $x^5=1$ with $x\neq 1$ then find value of $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3}$$


So my first observation was x is a non real fifth root of unity. Also $$x^5-1=(x-1)(1+x+x^2+x^3+x^4)=0$$
Thus $$1+x+x^2+x^3+x^4=0$$ I tried using this condition to simplify the above expression but nothing interesting simplified. Please note  i am looking for hints rather than complete solutions.
EDIT:I came to know its a duplicate ,but i feel that the answers given below are different from those in the original.
 [/INST] $$\frac{x}{1+x^2}+\frac{x^2}{1+x^4}+\frac{x^3}{1+x}+\frac{x^4}{1+x^3} =$$
$$=\frac{x}{1+x^2}\cdot\frac{x^4}{x^4}+\frac{x^2}{1+x^4}\cdot\frac{x^3}{x^3}+\frac{x^3}{1+x}\cdot \frac{x^2}{x^2}+\frac{x^4}{1+x^3}\cdot\frac{x}{x}= $$
(remember that $x^5=1$, so $x^6=x$ and $x^7=x^2$)
$$=\frac{1}{x^4+x}+\frac{1}{x^3+x^2}+\frac{1}{x^2+x^3}+\frac{1}{x+x^4}=
2\left(\frac{1}{x+x^4}+\frac{1}{x^2+x^3}\right)=$$
$$= 2\left(\frac{x^2+x^3+x+x^4}{(x+x^4)(x^2+x^3)} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x^6+x^7} \right) = 2\left(\frac{x+x^2+x^3+x^4}{x^3+x^4+x+x^2} \right) =2.
$$
</s>