Solve the following system of equations:

$$
\begin{gathered}
x+y+z=6 \\
x^{2}+y^{2}+z^{2}=14 \\
x^{3}+y^{3}+z^{3}=36 .
\end{gathered}
$$
The given equations can also be written as:

$$
(y+z)\left(y^{2}-y z+z^{2}\right)=36-x^{2}
$$

Subtracting (2) from the square of (1):

$$
2 y z+14-x^{2}=36-12 x+x^{2}
$$

from which

$$
y z=x^{2}-6 x+11
$$

Substituting (1), (2), and (4) into (3), performing the indicated operations, and rearranging the equation, we get:

$$
x^{3}-6 x^{2}+11 x-6=0
$$

If we substitute $(u-2)$ for $x$ in this equation, we obtain:

$$
u^{3}-u=0
$$

from which the values of $u$ are: $-1,0,1$, so the values of $x$ are: $1,2,3$.

Substituting the values of $x$ into (1) and (2), squaring (1), subtracting (2) from it, and finally dividing the equation by 2, we get the following systems of equations:

$$
\begin{array}{ccc}
y+z=5 & y+z=4 & y+z=3 \\
y z=6 & y z=3 & y z=2
\end{array}
$$

Solving these systems of equations, we obtain the roots of the equations:

$$
\begin{aligned}
& x \mid 1,1,2,2,3,3 \\
& y \mid 2,3,3,1,1,2 \\
& z \mid 3,2,1,3,2,1
\end{aligned}
$$

(Jenő Filkorn, Nyitra.)

The problem was solved by: Bayer B., Bertrám J., Burján K., Czank K., Fait F., Freibauer E., Kerekes T., Krausz B., Krisztián Gy., Kürth A., Lukhaub Gy., Lupsa Gy., Messik G., Perl Gy., Petrogalli G., Póka Gy., Sasvári G., Sasvári J., Stromfeld F., Szabó J.