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Question Number 70885 by naka3546 last updated on 09/Oct/19

$$\sqrt[3]{2}=a+\frac{1}{b+\frac{1}{c+\frac{1}{d+\dots }}}$$$$a,b,c,d\in {\mathbb{Z}}^{+}$$$${What}^{\prime }sb?$$

Answered by MJS last updated on 09/Oct/19

$$\mathrm{the}\mathrm{continued}\mathrm{fraction}\mathrm{of}\sqrt[3]{2}\mathrm{is}\mathrm{non}+\mathrm{periodic}$$$$\Rightarrow \mathrm{just}\mathrm{calculate}\mathrm{it}$$$$a=1$$$$b=\lfloor \frac{1}{\sqrt[3]{2}-1}\rfloor =\lfloor \sqrt[3]{4}+\sqrt[3]{2}+1\rfloor =3$$

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