Example 9 Let
$$
x^{3}-3 \sqrt{2} x^{2}+6 x-2 \sqrt{2}-8=0 \text {. }
$$

Then the value of $x^{5}-41 x^{2}+1$ is ( ).
(A) $13-\sqrt{2}$
(B) $-13+\sqrt{2}$
(C) -13
(D) 13
Solve the given conditions:
$$
\begin{array}{l}
(x-\sqrt{2})^{3}=8 \Rightarrow x-\sqrt{2}=2 \\
\Rightarrow(x-2)^{2}=2 \Rightarrow x^{2}-4 x+2=0 .
\end{array}
$$

Therefore, $x^{5}-41 x^{2}+1$
$$
\begin{array}{l}
=\left(x^{2}-4 x+2\right)\left(x^{3}+4 x^{2}+14 x+7\right)-13 \\
=-13 .
\end{array}
$$