## Task 10/87

Determine all triples $(x ; y ; z)$ of real numbers $x, y$, and $z$ that satisfy the system of equations:

$$
\begin{aligned}
x+y+z & =1 \\
x^{2}+y^{2}+z^{2} & =1 \\
x^{3}+y^{3}+z^{3} & =1
\end{aligned}
$$
Obviously, the triples $(1 ; 0 ; 0),(0 ; 1 ; 0)$ and $(0 ; 0 ; 1)$ are solutions of the system of equations (verified by substitution!). Due to (2), for $x ; y ; z \neq 1$, the inequalities $0 \leq x^{2} ; y^{2} ; z^{2}<1$ hold.

This implies $0 \leq|x| ;|y| ;|z|<1$.

From $a \leq|a|$ and $0 \leq|a|<1$, it follows that $a^{3} \leq\left|a^{3}\right|=a^{2}|a|<a^{2}$. Therefore, for $x ; y ; z \neq 1$, the inequality

$$
x^{3}+y^{3}+z^{3}<x^{2}+y^{2}+z^{2}=1
$$

holds.

Consequently, in this case, the equations

$$
x^{2}+y^{2}+z^{2}=1 \quad \text { and } \quad x^{3}+y^{3}+z^{3}=1
$$

are never satisfied simultaneously. The triples listed under 1. are thus the only solutions of the system of equations.