## Problem 4

Find all complex numbers $x, y, z$ which satisfy $x+y+z=x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}=3$.


## Solution

Answer: $1,1,1$.

We have $(x+y+z)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x)$, so $x y+y z+z x=3$.

$(x+y+z)^{3}=\left(x^{3}+y^{3}+z^{3}\right)+3\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+6 x y z$, so $8=\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x\right.$ $\left.+z x^{2}\right)+2 x y z$. But $(x+y+z)(x y+y z+z x)=\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+3 x y z$, so $x y z=-1$. Hence $x, y, z$ are the roots of the cubic $w^{3}-3 w^{2}+3 w-1=(w-1)^{3}$. Hence $x=y=z=1$.

