$$
\begin{array}{l}
\text { Example } 4 \text { Given } \frac{x-a-b}{c}+\frac{x-b-c}{a}+ \\
\frac{x-c-a}{b}=3 \text {, and } \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0 \text {. Then } x-a- \\
b-c=
\end{array}
$$

Example 4 Given $\frac{x-a-b}{c}+\frac{x-b-c}{a}+$ $\frac{x-c-a}{b}=3$, and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$. Then $x-a-$ $b-c=$
（8th Jinyun Cup Junior High School Mathematics Invitational Competition）
Solution: The given equation contains four letters $x, a, b, c$, and can certainly be treated as an equation in $x$, solving for $x$ and then finding the value, but this is a bit cumbersome. Instead, it is better to treat the expression to be evaluated, $x-a-b-c$, as the unknown. Thus, we get the following solution:
By moving terms on the right side of the equation, we get
$$
\begin{array}{l}
\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a} \\
+\frac{x-a-b-c}{b}=0, \\
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(x-a-b-c)=0 .
\end{array}
$$

Since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \neq 0$, we immediately get $x-a-b-c=0$.