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Question Number 170914 by mathlove last updated on 03/Jun/22

Commented by infinityaction last updated on 03/Jun/22

$$x=\frac{1}{{}^3\sqrt{2}-1}g.p.sum$$$$1+\frac{1}{x}={}^3\sqrt{2}$$$${(1+\frac{1}{x})}^3=2$$

Commented by cortano1 last updated on 04/Jun/22

$${(1+\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}.\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1})}^3$$$$={(1+\frac{\sqrt[3]{2}-1}{{\left(\sqrt[3]{2}\right)}^3-1})}^3$$$$={(1+\frac{\sqrt[3]{2}-1}{2-1})}^3={\left(\sqrt[3]{2}\right)}^3=2$$

Commented by mathlove last updated on 04/Jun/22

$$tanksforall$$

Answered by mahdipoor last updated on 03/Jun/22

$$({}^3\sqrt{4}{+}^3\sqrt{2}+1)({}^3\sqrt{2}-1)=1\Rightarrow 1/x{=}^3\sqrt{2}-1$$$${(1+1/x)}^3={(1{+}^3\sqrt{2}-1)}^3=2$$

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