Example 6. Given $a, b, c$ satisfy $a+b+c=1, a^{2}+$ $b^{2}+c^{2}=2, a^{3}+b^{3}+c^{3}=3$. Find the values of $a b c, a^{4}+b^{4}+c^{4}$.
(1991, Jiangsu Province Junior High School Mathematics Competition)
Let $S_{n}=a^{n}+b^{n}+c^{n}$.
$$
\begin{array}{l}
\because(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2(a b+a c+b c), \\
a+b+c=1, a^{2}+b^{2}+c^{2}=2 c \\
\therefore a b+a c+b c=-\frac{1}{2} .
\end{array}
$$

There is also the identity $x^{3}-(x+y+z) x^{2}+(x y$
$$
+y z+z x)+x y z=0 \text {. }
$$

From this, we get $a^{3}-a^{2}-\frac{1}{2} a-a b c=0$,
$$
\begin{array}{l}
b^{3}-b^{2}-\frac{1}{2} b-a b c=0 . \\
c^{3}-c^{2}-\frac{1}{2} c-a b c=0 .
\end{array}
$$
(1) $\times a^{n-2}+$ (2) $\times b^{n-3}+$ (3) $\times c^{n-3}$, we obtain the recurrence relation
$$
\begin{array}{l}
S_{n}=S_{n-1}+\frac{1}{2} S_{n-2}+(a b c) S_{n-3} . \\
\because S_{3}=3, S_{2}=2, S_{1}=1, \text { and } \\
S_{3}=S_{2}+\frac{1}{2} S_{1}+3 a b c, \\
\therefore 3=2+\frac{1}{2} \times 1+3 a b c, \\
\therefore a b c=\frac{1}{6} .
\end{array}
$$

Thus, $S_{4}=S_{3}+\frac{1}{2} S_{2}+(a b c) S_{1}$
$$
=3+\frac{1}{2} \times 2+\frac{1}{6} \times 1=\frac{25}{6} .
$$