Example 5 If real numbers $a, b, c$ satisfy $a+b+c=1, a^{2}+b^{2}+c^{2}=2, a^{3}+b^{3}+c^{3}=3$, find the values of $a b c$ and $a^{4}+b^{4}+c^{4}$.
Analysis Let $a, b, c$ be the roots of the polynomial $f(t)=t^{3}-\sigma_{1} t^{2}+\sigma_{2} t-\sigma_{3}$. Denote $s_{k}=a^{k}+b^{k}+c^{k} (k=1,2,3,4)$. By Newton's formulas, we can find $\sigma_{1}, \sigma_{2}, \sigma_{3}$, and then solve the problem.

Solution Let $a, b, c$ be the roots of the polynomial $f(t)=t^{3}-\sigma_{1} t^{2}+\sigma_{2} t-\sigma_{3}$, and denote $s_{k}=a^{k}+b^{k}+c^{k}$, $k=1,2,3,4$. Then, by Newton's formulas and the given conditions, we have $s_{1}=1=\sigma_{1}, s_{2}=2=\sigma_{1} s_{1}-2 \sigma_{2}, s_{3}=3=\sigma_{1} s-\sigma_{2} s+3 \sigma_{3}$,
Solving these, we get $\sigma_{1}=1, \sigma_{2}=-\frac{1}{2}, \sigma_{3}=\frac{1}{6}$, thus $a b c=\sigma_{3}=\frac{1}{6}$, and $s_{4}=\sigma_{1} s_{3}-\sigma_{2} s_{2}+\sigma_{3} s_{1}=\frac{25}{6}$, i.e., $a^{4}+b^{4}+c^{4}=\frac{25}{6}$.