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.

\begin{align}
\left( \sum_{cyc} x \right) \left( \sum_{\text{cyc}} x^2
\right) &= \left( \sum_{\text{cyc}} x^3 \right) = 1 \\
\sum_{cyc} x^2y + \sum_{cyc} xy^2 &= 0
\end{align}
In the above equation, subtract $x^3+y^3+z^3$ from both sides.
\begin{align}
(x+y+z)^3 &= 1 \\
\sum_{cyc} x^3 + 3\sum_{cyc} x^2y + 3\sum_{cyc} xy^2 + 6xyz &= 1 \\
1 + 0 + 0 + 6xyz &= 1
\end{align}
Hence $xyz=0$.
