Example 8 Solve the equation
$$
\begin{array}{l}
\quad \frac{b+c}{x-a}+\frac{c+a}{x-b}+\frac{a+b}{x-c}=\frac{3(a+b+c)}{x} . \\
(a>0, b>0, c>0)
\end{array}
$$
$$
\begin{array}{l}
\left(\frac{b+c}{x-a}-\frac{a+b+c}{x}\right)+\left(\frac{c+a}{x-b}-\frac{a+b+c}{x}\right) \\
+\left(\frac{a+b}{x-c}-\frac{a+b+c}{x}\right)=0 . \\
\because \frac{b+c}{x-a}-\frac{a+b+c}{x} \\
\quad=\left(\frac{b+c}{x-a}-1\right)+\left(1-\frac{a+b+c}{x}\right) \\
\quad=\frac{a+b+c-x}{x-a}-\frac{a+b+c}{x} \\
\quad=(a+b+c-x)\left(\frac{1}{x-a}-\frac{1}{x}\right)
\end{array}
$$

Since in the original equation, $a$, $b$, and $c$ are cyclically symmetric, the original equation can be transformed into
$$
\begin{array}{l}
(a+b+c-x)\left(\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}-\frac{3}{x}\right) \\
=0 . \\
\therefore \quad x_{1}=a+b+c
\end{array}
$$

or $\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}-\frac{3}{x}=0$.
By eliminating the denominators and rearranging from $\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}-\frac{3}{x}=0$,

we get
$$
\begin{array}{c}
(a+b+c) x^{2}-2(a b+b c+c a) x+3 a b c \\
=0 \\
\therefore \quad x_{2,3}= \\
\frac{(a b+b c+c a) \pm \sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}-a b c(a+b+c)}}{a+b+c} .
\end{array}
$$