By Theorem 4, let $\xi_{0}=\sqrt[3]{2}$.
Therefore,
$$\sqrt[3]{2}=\left\langle 1,3,1,5,1,1, \xi_{6}\right\rangle .$$
$$\begin{array}{l}
a_{0}=\left[\xi_{0}\right]=1, \quad \xi_{1}=\left(\xi_{0}-a_{0}\right)^{-1}=(\sqrt[3]{2}-1)^{-1} \\
=\sqrt[3]{4}+\sqrt[3]{2}+1, \\
a_{1}=\left[\xi_{1}\right]=3, \quad \xi_{2}=\left(\xi_{1}-3\right)^{-1}=(\sqrt[3]{4}+\sqrt[3]{2}-2)^{-1} \\
=\frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{\sqrt[3]{2}+2} \\
a_{2}=\left[\xi_{2}\right]=1, \quad \xi_{3}=\left(\xi_{2}-1\right)^{-1}=\frac{\sqrt[3]{2}+2}{\sqrt[3]{4}-1} \\
=\frac{4 \sqrt[3]{4}+5 \sqrt[3]{2}+4}{3} \\
a_{3}=\left[\xi_{3}\right]=5, \quad \xi_{4}=\left(\xi_{3}-5\right)^{-1}=\frac{3}{4 \sqrt[3]{4}+5 \sqrt[3]{2}-11}, \\
a_{4}=\left[\xi_{4}\right]=1, \quad \xi_{5}=\left(\xi_{4}-1\right)^{-1}=\frac{4 \sqrt[3]{4}+5 \sqrt[3]{2}-11}{14-5 \sqrt[3]{2}-4 \sqrt[3]{4}}, \\
a_{5}=\left[\xi_{5}\right]=1, \quad \xi_{6}=\left(\xi_{5}-1\right)^{-1}=\frac{14-5 \sqrt[3]{2}-4 \sqrt[3]{4}}{8 \sqrt[3]{4}+10 \sqrt[3]{2}-25}
\end{array}$$
Example 5 Find the first six digits of the infinite simple continued fraction for $\sqrt[3]{2}$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.