4. If the real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text {, }
$$

and $x-z=33$,
then the value of the algebraic expression $x-2 y+z$ is
4. $\pm 31$.

Let $a=\sqrt[5]{x-y}, b=\sqrt[5]{y-z}$. Then
$$
\begin{array}{l}
a+b=3, \\
a^{5}+b^{5}=33 .
\end{array}
$$
(2) $\div$ (1) gives
$$
\begin{array}{l}
a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4}=11 \\
\Rightarrow\left(a^{2}+b^{2}\right)^{2}-a b\left(a^{2}+b^{2}\right)-a^{2} b^{2}=11 \\
\Rightarrow\left[(a+b)^{2}-2 a b\right]^{2}- \\
\quad a b\left[(a+b)^{2}-2 a b\right]-a^{2} b^{2}=11 \\
\Rightarrow(9-2 a b)^{2}-a b(9-2 a b)-a^{2} b^{2}=11 \\
\Rightarrow 5 a^{2} b^{2}-45 a b+70=0 \\
\Rightarrow a b=2 \text { or } 7 .
\end{array}
$$

Solving (1) and (3) simultaneously, we get
$$
\left\{\begin{array} { l } 
{ a = 1 , } \\
{ b = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
a=2, \\
b=1 .
\end{array}\right.\right.
$$

Therefore, $x-2 y+z=a^{5}-b^{5}= \pm 31$.