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.

a nice problem! squaring the first equation we obtain
$$xy+xz+yz=0$$
raise to the power $3$ the last equation we have
$$x^2(y+z)+y^2(x+z)+z^2(x+y)+xyz=0$$
and this can be written as
$$x^2(1-x)+y^2(1-y)+z^2(1-z)+xyz=0$$
and this is
$$x^2+y^2+z^2-x^3-y^3-z^3+xyz=0$$ therefore $xyz=0$
