## Problem 2.

Solve, in the set of real numbers, the system of equations

$$
\left.\begin{array}{l}
y^{3}-6 x^{2}+12 x-8=0 \\
z^{3}-6 y^{2}+12 y-8=0 \\
x^{3}-6 z^{2}+12 z-8=0
\end{array}\right\}
$$
Solution. Since $3 t^{2}-6 t+4 \geq 0$ for all $t \in \mathbb{R}$, any solution $\left(x_{0}, y_{0}, z_{0}\right)$ of the system satisfies $x_{0}^{3}>0, y_{0}^{3}>0, z_{0}^{3}>0$. That is, $x_{0}, y_{0}, z_{0}$ are positive numbers.

Moreover, adding the three equations results in

$$
\left(x^{3}-6 x^{2}+12 x-8\right)+\left(y^{3}-6 y^{2}+12 y-8\right)+\left(z^{3}-6 z^{2}+12 z-8\right)=0
$$

or equivalently,

$$
(x-2)^{3}+(y-2)^{3}+(z-2)^{3}=0
$$

Now we will distinguish two cases: (a) $x_{0} \geq 2$ and (b) $0 < x_{0} < 2$. If $x_{0} \geq 2$, then $(x_{0}-2)^{3} \geq 0$. Since $(x_{0}-2)^{3}+(y_{0}-2)^{3}+(z_{0}-2)^{3}=0$, it follows that $(y_{0}-2)^{3}+(z_{0}-2)^{3} \leq 0$. If $y_{0} \geq 2$, then $(y_{0}-2)^{3} \geq 0$, which implies $(z_{0}-2)^{3} \leq 0$. Since $z_{0} > 0$, if $z_{0} < 2$, then $(z_{0}-2)^{3} < 0$, which contradicts the equality (1). Similarly, it follows that $y_{0} < 2$, which contradicts the equality (1). Consequently, the only real solution of the system is $(2,2,2)$ and we are done.