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Number theory Difficulty 6.8 National olympiad Find the answer

Example 5 Find the first six digits of the infinite simple continued fraction for 23\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.

A number or a short expression. Spacing and $ signs are ignored.

Solution

By Theorem 4, let ξ0=23\xi_{0}=\sqrt[3]{2}.
Therefore,
23=⟨1,3,1,5,1,1,ξ6⟩.\sqrt[3]{2}=\left\langle 1,3,1,5,1,1, \xi_{6}\right\rangle .
a0=[ξ0]=1,ξ1=(ξ0−a0)−1=(23−1)−1=43+23+1,a1=[ξ1]=3,ξ2=(ξ1−3)−1=(43+23−2)−1=43+23+123+2a2=[ξ2]=1,ξ3=(ξ2−1)−1=23+243−1=443+523+43a3=[ξ3]=5,ξ4=(ξ3−5)−1=3443+523−11,a4=[ξ4]=1,ξ5=(ξ4−1)−1=443+523−1114−523−443,a5=[ξ5]=1,ξ6=(ξ5−1)−1=14−523−443843+1023−25\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}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.