Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things.

For $x:0\neq|x|<1$ we have $x^3<x^2$. Therefore if we have 3 numbers $x_1^2+x_2^2+x_3^2=1$ and $x_1x_2x_3\neq0$ we have that $x_i\neq0$ $\forall i$. Therefore for every $i$ we have the strict inequallity $x_i^3<x_i^2$ therefore $x_1^3+x_2^3+x_3^3<1$. Therefore one of the $x_i$ is $0$ .
