Eliminate $x,y,z$

$x+y+z=0$, $x^2+y^2+z^2=a^2$, $x^3+y^3+z^3=b^5$, $x^5+y^5+z^5=c^5$ to find the relation between a b and c

Firstly xy+yz+zx= (-a^2)/2. And if x+y+z=0 then x^3+y^3+z^3= 3xyz, now from this xyz=b^5/3 now multiply 2nd and 3rd equation and after manipulating it we can easily get out 5(a^2.b^5)=6(c^5)