Solve the following system of equations:

$$
\begin{aligned}
x+y+z & =3 \\
x^{2}+y^{2}+z^{2} & =7 \\
x^{3}+y^{3}+z^{3} & =15
\end{aligned}
$$
Solution. The system of equations remains unchanged under the exchange of any two variables, so if we find one solution, all its possible permutations will also be solutions. First, we calculate the values of $x y+y z+z x$ and $x y z$. Based on notable identities, on the one hand:

$$
9=(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x=7+2(x y+y z+z x)
$$

from which it follows that $x y+y z+z x=1$. On the other hand, by expanding the cube of a trinomial and similar transformations:

$$
\begin{aligned}
27 & =(x+y+z)^{3}= \\
& =x^{3}+y^{3}+z^{3}+3\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+6 x y z= \\
& =15+3\left(x^{2} y+x y^{2}+x y z+y^{2} z+y z^{2}+x y z+z^{2} x+z x^{2}+x y z\right)-3 x y z= \\
& =15+3(x+y+z)(x y+y z+z x)-3 x y z=15+3 \cdot 3-3 x y z
\end{aligned}
$$

By comparing the two sides, we get that $x y z=-1$. It is also clear from this that none of the roots can be zero. After some algebraic transformations, we arrive at a cubic equation in the variable $x$. The sum can be expressed as $y+z$:

$$
y+z=3-x
$$

From the product, we can immediately express $y z$ in terms of $x$: $y z=-\frac{1}{x}$. Using these two facts, only one variable remains in the equation:

$$
1=x y+y z+z x=x(y+z)+y z=x(3-x)-\frac{1}{x}
$$

The resulting equation is:

$$
1=3 x-x^{2}-\frac{1}{x}
$$

and after rearrangement, $x^{3}-3 x^{2}+x+1=0$. This equation has the solution $x=1$, so the factor $(x-1)$ can be factored out:

$$
(x-1)\left(x^{2}-2 x-1\right)=0
$$

The other two roots of the equation are $x=1+\sqrt{2}$ and $x=1-\sqrt{2}$. Due to symmetry, all permutations of these three numbers give all solutions of the original system of equations. Therefore, the system of equations has six solutions. It can be verified by substitution that the obtained values are indeed roots of the system of equations.

Remarks. 1. In solving the system of equations, we obtained the values of $x+y+z$, $x y+y z+z x$, and $x y z$, which are the Viète's formulas of a cubic equation. Thus, all solutions of the system of equations can be obtained as the roots of the equation $u^{3}-3 u^{2}+u+1=0$. This exactly matches the solution described earlier.

2. Contestants who did not detail the solution of the cubic equation or did not verify the obtained roots did not receive full marks. This is the cause of the relatively large number of incomplete solutions.