Example 8 Given real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text{, and } x-z=33 \text{. }
$$

Find the value of the algebraic expression $x-2 y+z$.
Let $\sqrt[5]{x-y}=x_{1}, \sqrt[5]{y-z}=x_{2}$. Then $x_{1}+x_{2}=3, x_{1}^{5}+x_{2}^{5}=33$.
Let $x_{1} x_{2}=\lambda$. Then $x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+\lambda=0$.
Let $x_{1}^{n}+x_{2}^{n}=a_{n}$ ( $n$ is a positive integer $)$. Then
$$
a_{n}-3 a_{n-1}+\lambda a_{n-2}=0 \text {, }
$$

i.e.,
$$
a_{n}=3 a_{n-1}-\lambda a_{n-2}(n \geqslant 3) .
$$

From $x_{1}+x_{2}=3, x_{1} x_{2}=\lambda$
$$
\begin{array}{l}
\Rightarrow a_{1}=3, a_{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=9-2 \lambda \\
\Rightarrow a_{3}=3(9-2 \lambda)-3 \lambda=27-9 \lambda \\
\Rightarrow a_{4}=3(27-9 \lambda)-\lambda(9-2 \lambda) \\
\quad=81-36 \lambda+2 \lambda^{2} \\
\Rightarrow a_{5}=3\left(81-36 \lambda+2 \lambda^{2}\right)-\lambda(27-9 \lambda)=33 \\
\Rightarrow 81-36 \lambda+2 \lambda^{2}-\lambda(9-3 \lambda)=11 \\
\Rightarrow \lambda^{2}-9 \lambda+14=0
\end{array}
$$
$$
\Rightarrow \lambda=2 \text { or } 7 \text {. }
$$

From $x^{2}-3 x+\lambda=0$ having two real roots, we get
$$
\Delta=(-3)^{2}-4 \lambda \geqslant 0 \Rightarrow \lambda \leqslant \frac{9}{4}
$$
$\Rightarrow \lambda=2 \Rightarrow x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+2=0$
$$
\Rightarrow\left\{\begin{array} { l } 
{ x _ { 1 } = 1 , } \\
{ x _ { 2 } = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=2, \\
x_{2}=1 .
\end{array}\right.\right.
$$

Since $x-2 y+z=x_{1}^{5}-x_{2}^{5}$, we have
$$
1^{5}-2^{5}=-31 \text { or } 2^{5}-1^{5}=31
$$

This is the solution.