Question 4 Let $a, b, c \in \mathbf{C}$, satisfying
$$
\left\{\begin{array}{l}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=9, \\
\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}=32, \\
\frac{a^{3}}{b+c}+\frac{b^{3}}{c+a}+\frac{c^{3}}{a+b}=122 .
\end{array}\right.
$$

Find the value of $a b c$.
[4]
(2017, Carnegie Mellon University Informatics and Mathematics Competition)
【Analysis】Observe the relationship between equations (1) and (2). Try multiplying both sides of equation (1) by \(a+b+c\), we get
$$
\begin{array}{l}
\frac{a^{2}}{b+c}+\frac{a(b+c)}{b+c}+\frac{b^{2}}{c+a}+\frac{b(c+a)}{c+a}+ \\
\frac{c^{2}}{a+b}+\frac{c(a+b)}{a+b} \\
=9(a+b+c) \\
\Rightarrow 32+a+b+c=9(a+b+c) \\
\Rightarrow a+b+c=4 .
\end{array}
$$

Similarly, multiplying both sides of equation (2) by \(a+b+c\) gives
$$
a^{2}+b^{2}+c^{2}=6 \text {. }
$$

Given that the denominators of the three equations are the same, try to find
$$
I=\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \text {. }
$$

Multiplying both sides by \(a+b+c\) gives
$$
4 I=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \text {. }
$$

Substituting equation (1) into the above equation, we get \(I=3\).
Since \(a+b+c=4\), the above equation can be rewritten as
$$
\frac{1}{4-a}+\frac{1}{4-b}+\frac{1}{4-c}=3 \text {. }
$$

Considering \(a, b, c\) as the roots of the cubic polynomial \(x^{3}-4 x^{2}+a_{1} x+a_{0}\), we can apply the proposition.
From equations (3) and (4), we get \(a b+b c+a c=5\).
Applying Vieta's formulas, we find \(a_{1}=5\).
Notice that,
$$
\frac{1}{4-a}+\frac{1}{4-b}+\frac{1}{4-c}=\frac{P^{\prime}(4)}{P(4)}=3 \text {. }
$$

Let \(P(x)=x^{3}-4 x^{2}+5 x+a_{0}\). Then
$$
\begin{array}{l}
P^{\prime}(x)=3 x^{2}-8 x+5, \\
P(4)=20+a_{0}, P^{\prime}(4)=21 .
\end{array}
$$

Substituting into equation (6), we solve for \(a_{0}=-13\).
We need to find \(a b c\), which by Vieta's formulas is \(a b c=13\).
It is worth noting that: multiplying both sides of equation (5) by
$$
(4-a)(4-b)(4-c),
$$