Question

Eliminate $x,y,z$ from the equations $x+y+z=0,x^2+y^2+z^2=a^2,x^3+y^3+z^3=b^3,x^5+y^5+z^5=c^5$.

Answer 1

$x+y+z=0$ ..... $\left(i\right)$

$x^2+y^2+z^2=a^2$..... $\left(ii\right)$

$x^3+y^3+z^3=b^3$..... $\left(iii\right)$

$x^5+y^5+z^5=c^5$ ..... $\left(iv\right)$

Consider

$\Rightarrow$$(x^2+y^2+z^2)(x^3+y^3+z^3)=a^2{b}^3$

$\Rightarrow$$(x^5+y^5+z^5+x^2{y}^2(x+y)+x^2{z}^2(x+z)+y^2{z}^2(y+z)=a^2{b}^3$

$\Rightarrow$$c^5+x^2{y}^2(-z)+x^2{z}^2(-y)+y^2{z}^2(-x)=a^2{b}^3$

$\Rightarrow$$c^5-xyz(xy+yz+zx)=a^2{b}^3$

$\Rightarrow$$c^5-\frac{b^3}{3}\left(\frac{-a^2}{2}\right)=a^2{b}^3$

$\Rightarrow$$c^5+\left(\frac{a^2{b}^3}{6}\right)=a^2{b}^3$

$\Rightarrow$$c^5=a^2{b}^3-\frac{a^2{b}^3}{6}$

$\Rightarrow$$5a^2{b}^3=6c^5$