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Question Number 138366 by KwesiDerek last updated on 12/Apr/21

$$\mathbf{If}{\mathbf{x}}^2+{\mathbf{x}}^{-2}=\sqrt{2+\sqrt{2+\sqrt{2}}}$$$${\mathbf{x}}^{16}+{\mathbf{x}}^{-16}=?$$$$\mathbf{Any}\mathbf{help}$$

Answered by TheSupreme last updated on 14/Apr/21

$$x^4+1-\sqrt{2+\sqrt{2+\sqrt{2}}}{x}^2=0$$$$x^2=tandsolvefort$$$$thanevaluate$$$$t^4+\frac{1}{t^4}$$

Answered by Ñï= last updated on 14/Apr/21

$$x^2+x^{-2}=\sqrt{{(x^2+x^{-2})}^2}=\sqrt{2+x^4+x^{-4}}$$$$=\sqrt{2+\sqrt{{(x^4+x^{-4})}^2}}=\sqrt{2+\sqrt{2+x^8+x^{-8}}}$$$$=\sqrt{2+\sqrt{2+\sqrt{{(x^8+x^{-8})}^2}}}=\sqrt{2+\sqrt{2+\sqrt{2+x^{16}+x^{-16}}}}$$$$\Rightarrow x^{16}+x^{-16}=0$$

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