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.

$$x^3+y^3+z^3-3xyz=(x+y+z)\{(x+y+z)^2-3(xy+yz+zx)\}$$
Now $$2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=?$$
