Marinchoo wrote:
Find $A=x^5+y^5+z^5$ if $x+y+z=1$, $x^2+y^2+z^2=2$ and $x^3+y^3+z^3=3$.
$xy+yz+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}2=-\frac 12$
$x^3+y^3+z^3=(x+y+z)(x^3+y^3+z^3)-(xy+yz+zx)(x+y+z)+3xyz$ and so $xyz=\frac 16$
So $x,y,z$ are roots of $x^3-x^2-\frac x2-\frac 16=0$ (one real, the two other complex)
Let $S_n=x^n+y^n+z^n$
$S_1=1,S_2=2,S_3=3$ and $S_{n+3}=S_{n+2}+\frac 12S_{n+1}+\frac 16S_n$
So $S_4=\frac {25}6$ and $S_5=\boxed{x^5+y^5+z^5=6}$