<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] Answer :
((x / (1+x² ))+((x³/ (1+x))=((x+x² +x³ +x⁵) / ((1+x)(1+x²) ))=((x+x² +x³ +x⁵) / (x+x² +x³ +1))=1
Because x^5=1
$((x² / (1+x⁴))+((x⁴/ (1+x³ )) = ((x² +x⁵+x⁴+x⁸) / (1+x³ +x⁴+x^7)) =((x² +1+x⁴+x³) / (1+x³+x⁴+x²)) =1$
Because $x^5 =1$
$\frac{x}{1+x^2 }+\frac{x^3 }{1+x} +\frac{x^2 }{1 +x^4}+\frac{x^4}{1+x^3 }=\frac{x+x^2 +x^3 +x^5}{(1+x)(1+x^2) }+\frac{x^2 +x^5+x^4+x^8}{(1+x^4)(1+x^3) }=\frac{x+x^2 +x^3 +x^5}{x+x^2 +x^3 +1}+\frac{x^2 +1+x^4+x^3 }{1+x^3 +x^4+x^2} =2$
Because $x^5 =1\Rightarrow x^7=x^2$ and $x ^8 = x^3$
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