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Question Number 54526 by Gulay last updated on 05/Feb/19

Commented by Gulay last updated on 05/Feb/19

$$\mathrm{sir}\mathrm{plz}\mathrm{help}\mathrm{me}$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19

$$solvebyreasoning...$$$$\sqrt[3]{16+8\sqrt{5}}since$$$$1)look\sqrt{5}ispresent$$$$2)lookallare+vesign$$$$so8\sqrt{5}=3\sqrt{5}+5\sqrt{5}=3\times 1^2\times \sqrt{5}+{\left(\sqrt{5}\right)}^3$$$$16=1+3\times 1\times {\left(\sqrt{5}\right)}^2=1+15$$$$$$$$so16+8\sqrt{5}$$$$=1^3+3\times 1^2\times \sqrt{5}+3\times 1\times {\left(\sqrt{5}\right)}^2+{\left(\sqrt{5}\right)}^3$$$$={(1+\sqrt{5})}^3$$$$hence\sqrt[3]{16+8\sqrt{5}}=1+\sqrt{5}$$$$now\sqrt{5+\sqrt{5}+\sqrt[3]{16+8\sqrt{5}}}$$$$=\sqrt{5+\sqrt{5}+1+\sqrt{5}}$$$$=\sqrt{6+2\sqrt{5}}$$$$=\sqrt{5+1+2\sqrt{5}}$$$$=\sqrt{{\left(\sqrt{5}\right)}^2+2\times \sqrt{5}\times 1+{\left(1\right)}^2}$$$$=\sqrt{{(\sqrt{5}+1)}^2}$$$$=\sqrt{5}+1$$

Commented by Gulay last updated on 05/Feb/19

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}$$

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