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Question Number 126109 by mnjuly1970 last updated on 17/Dec/20

$$....nicecalculus...$$$$verifythat::\mathrm{A}=\frac{\sqrt[3]{2+\sqrt{5}}}{1+\sqrt{5}}is$$$$arationalnumber...$$

Answered by som(math1967) last updated on 17/Dec/20

$$\frac{\sqrt[3]{8(2+\sqrt{5)}}}{(1+\sqrt{5})\times 2}$$$$=\frac{\sqrt[3]{16+8\sqrt{5}}}{2(1+\sqrt{5})}$$$$=\frac{\sqrt[3]{1^3+3\times 1^2\times \sqrt{5}+3\times 1\times {\left(\sqrt{5}\right)}^2+{\left(\sqrt{5}\right)}^3}}{2(1+\sqrt{5})}$$$$=\frac{\sqrt[3]{{(1+\sqrt{5})}^3}}{2(1+\sqrt{5})}=\frac{1}{2}isarational$$$$number.$$

Commented by mnjuly1970 last updated on 17/Dec/20

$$thanksalot$$

Answered by Dwaipayan Shikari last updated on 17/Dec/20

$$A=\frac{\sqrt[3]{1+t}}{t}=\sqrt{\frac{1}{t^3}+\frac{1}{t^2}}=\sqrt[3]{\frac{1}{{(1+\sqrt{5})}^3}+\frac{1}{{(1+\sqrt{5})}^2}}$$$$=\sqrt[3]{{\left(\frac{\sqrt{5}-1}{4}\right)}^2+{\left(\frac{\sqrt{5}-1}{4}\right)}^3}=\sqrt[3]{\frac{5\sqrt{5}-1-3\sqrt{5}(\sqrt{5}-1)}{64}+\frac{24-8\sqrt{5}}{64}}$$$$=\sqrt[3]{\frac{5\sqrt{5}-1-15+3\sqrt{5}+24-8\sqrt{5}}{64}}=\frac{1}{2}$$

Commented by mnjuly1970 last updated on 17/Dec/20

$$grateful...$$

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