Example 2. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=6 \\
x^{2}+y^{2}+z^{2}=14 \\
x^{3}+y^{3}+z^{3}=36
\end{array}\right.
$$
$$
\begin{array}{l}
\text { Sol } \quad \text { we have } x y+y z+z x=\frac{1}{2}\left[(x+y+z)^{2}\right. \\
\left.-\left(x^{2}+y^{2}+z^{2}\right)\right]=\frac{1}{2}(36-14)=11, \\
x y z=\frac{1}{3}\left\{x^{3}+y^{3}+z^{3}-(x+y+z)\right. \\
\text { - } \left.\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x)\right]\right\} \\
=6 .
\end{array}
$$

It is clear that $x, y, z$ are the roots of the equation
$$
u^{3}-6 u^{2}+11 u-6=0
$$

It is easy to see that the equation has roots $1, 2, 3$, thus we can obtain 6 solutions $(x, y, z)=(1,2,3),(1,3,2)$, etc.