Example 6 If the three roots of $x^{3}+a x^{2}+b x+c=0$ are $a, b, c$, and $a, b, c$ are rational numbers not all zero. Find $a, b, c$.
(2005, Shanghai Jiao Tong University Independent Recruitment Examination)
Solve by Vieta's formulas:
$$
\left\{\begin{array}{l}
a+b+c=-a, \\
a b+b c+c a=b, \\
a b c=-c .
\end{array}\right.
$$

From equation (1), we know $c=0$ or $a b=-1$.
(1) If $c=0$, then
$$
\left\{\begin{array}{l}
2 a + b = 0, \\
a b = b
\end{array} \Rightarrow \left\{\begin{array}{l}
a=1, \\
b=-2 .
\end{array}\right.\right.
$$
(2) If $a b=-1$, then
$$
\left\{\begin{array}{l}
2 a-\frac{1}{a}+c=0, \\
-1+a c-\frac{1}{a} \cdot c=-\frac{1}{a} .
\end{array}\right.
$$

Substituting equation (2) into equation (3) yields
$$
(a-1)\left(2 a^{3}+2 a^{2}-1\right)=0 \text {. }
$$

If $a=1$, then $b=-1, c=-1$.
$$
\text { If } 2 a^{3}+2 a^{2}-1=0 \text {, }
$$

Since $a$ is a rational number, let $a=\frac{p}{q}(p, q$ are coprime positive integers, and $q \geqslant 1$). Then
$$
\frac{2 p^{3}}{q^{3}}+\frac{2 p^{2}}{q^{2}}=1 \text {, }
$$

which simplifies to $2 p^{2}(p+q)=q^{3}$.
Since $p, q$ are coprime, $q^{3}$ must divide 2.
Thus, $q=1$.
Therefore, $a$ is an integer.
However, the left side of equation (4) is odd, while the right side is even, leading to a contradiction.
In summary, $(a, b, c)=(1,-2,0),(1,-1,-1)$.