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.

We have $$(x+y+z)(x^2-xy+y^2+z^2)=x^3+y^3+z^3+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$x^2-xy+y^2+z^2=1+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$-xy=xz(x+z)+yz(y+z)-xyz$$ giving $$xyz=xy+xz-xyz+yz-xyz\implies 3xyz=xy+xz+yz$$ Now $$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\implies xy=xz+yz=0$$ Hence $$\boxed{3xyz=0\implies xyz=0.}$$
