Example 3. Solve the equation: $\sqrt[5]{171-x}+\sqrt[5]{104+x}=5$.


Let $u=\sqrt[5]{171-x}, v=\sqrt[5]{104+x}$, then
$$
\left\{\begin{array}{l}
u+v=5 \\
u^{5}+v^{5}=275
\end{array}\right.
$$

From (2) we get $u^{5}+v^{5}=(u+n)\left(u^{2}+v^{4}\right)$
$$
\begin{array}{l}
-u v\left(u^{5}+v^{3}\right) \\
=\cdots=(u+v)^{5}-5 u v(u+v)^{3} \\
-1_{2}^{2} v^{2}(u+v) \\
=5^{5}-5^{4} u v-5^{2} v^{2} v^{2}=275 .
\end{array}
$$

Rearranging gives: $(u v)^{2}-25 u v+114=0$.
Solving it, we get $u v=6$ or $u v=19$.
$$
\text { Therefore, }\left\{\begin{array} { l } 
{ u + v = 5 , } \\
{ u v = 6 }
\end{array} \text { or } \left\{\begin{array}{l}
u+v=5, \\
u v=19 .
\end{array}\right.\right.
$$
(No solution in $\mathbb{R}$).
$$
\therefore\left\{\begin{array} { c } 
{ u = 2 } \\
{ v = 3 }
\end{array} \text { or } \left\{\begin{array}{l}
u=3, \\
v=2 .
\end{array}\right.\right.
$$

Substituting the values of $u$ and $v$ into the solution, we get $x_{1}=139$, $x_{2}=-72$, both of which are roots of the original equation after verification.