Example 5 Let $a, b, c$ be three of the four roots of the polynomial $x^{4}-a x^{3}-b x+c$. Find all such triples $a, b, c$.
(Former Czechoslovakia Olympiad Problem)
Let $c=0$, then $c$ is a root of the equation $x^{4}-a x^{3}-b x=0$. Therefore, what remains is to find all polynomials of the form $x^{3}-a x^{2}-b$ such that $a$ and $b$ are its roots. If we denote the third root of this polynomial by $d$, then by Vieta's formulas, we have $a+b+d=a$, which implies $d=-b$. Next,
$$
a b+a d+b d=a b-b(a+b)=-b^{2}=0,
$$

which means $b=0$. Finally, it can be verified that all triples of the form $(a, 0,0)$ satisfy the conditions of the problem. Now, let $c \neq 0$, then the polynomial $x^{4}-a x^{3}-b x+c$ has 4 roots $a, b, c, d$, none of which are 0. By Vieta's formulas, we have $a+b+c+d=a$, which implies $d=-(b+c)$.
Next, $a b+a c+b c+a d+b d+c d$
$$
\begin{array}{l}
=a b+a c+b c-(a+b+c)(b+c) \\
=-b^{2}-b c-c^{2} \\
=0 .
\end{array}
$$
$$
\begin{array}{l}
\text { Also, } a b c+a b d+a c d+b c d \\
= a b c-(a b+a c+b c)(b+c)=-a\left(b^{2}+b c+c^{2}\right)-b^{2} c-b c^{2} \\
=-b c(b+c)=b, \\
\text { which implies }-c(b+c)=1, c^{2}+b c+1=0 .
\end{array}
$$

Finally, $a b c d=-a b c(b+c)=a b=c$,
which implies $a=\frac{c}{b}$.
Therefore, $a, b, c$ can only have four sets:
$$
\begin{array}{l}
b_{1,2}=1, c_{1,2}=\frac{-1 \pm \mathrm{i} \sqrt{3}}{2}, a_{1,2}=\frac{-1 \pm \mathrm{i} \sqrt{3}}{2} ; \\
b_{3,4}=-1, c_{3,4}=\frac{1 \pm \mathrm{i} \sqrt{3}}{2}, a_{3,4}=\frac{-1 \mp \mathrm{i} \sqrt{3}}{2} .
\end{array}
$$

Each set satisfies the conditions of the problem. Therefore, the triples $(a, b, c)$ are either of the form $(a, 0,0)$, where $a$ is any complex number; or one of the following triples:
$$
\begin{array}{l}
\left(\frac{-1+\mathrm{i} \sqrt{3}}{2}, 1, \frac{-1+\mathrm{i} \sqrt{3}}{2}\right),\left(\frac{-1-\mathrm{i} \sqrt{3}}{2}, 1, \frac{-1-\mathrm{i} \sqrt{3}}{2}\right), \\
\left(\frac{-1-\mathrm{i} \sqrt{3}}{2},-1, \frac{1+\mathrm{i} \sqrt{3}}{2}\right),\left(\frac{-1+\mathrm{i} \sqrt{3}}{2},-1, \frac{1-\mathrm{i} \sqrt{3}}{2}\right) .
\end{array}
$$