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Question Number 143027 by mathlove last updated on 09/Jun/21

Commented by Rasheed.Sindhi last updated on 09/Jun/21

$$x+\frac{1}{x}=-1\Rightarrow x=\omega ,{\omega }^2$$$$Whenx=\omega$$$$x^{123456789}+\frac{1}{x^{123456789}}$$$$={\omega }^{123456789}+\frac{1}{{\omega }^{123456789}}$$$$={\omega }^{3\times 41152263}+\frac{1}{{\omega }^{3\times 41152263}}$$$$={\left({\omega }^3\right)}^{41152263}+\frac{1}{{\left({\omega }^3\right)}^{41152263}}$$$$={\left(1\right)}^{41152263}+\frac{1}{{\left(1\right)}^{41152263}}=2$$$$Whenx={\omega }^2$$$$Sameresult$$

Answered by mathmax by abdo last updated on 09/Jun/21

$$\mathrm{x}+\frac{1}{\mathrm{x}}=-1\Rightarrow \frac{{\mathrm{x}}^2+1}{\mathrm{x}}=-1\Rightarrow {\mathrm{x}}^2+1=-\mathrm{x}\Rightarrow {\mathrm{x}}^2+\mathrm{x}+1=0$$$$\mathrm{\Delta }=-3\Rightarrow {\mathrm{x}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}}\mathrm{and}{\mathrm{x}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-\frac{\mathrm{i}2\pi }{3}}$$$$\mathrm{ifx}={\mathrm{x}}_1\Rightarrow {\mathrm{x}}^{\mathrm{n}}+\frac{1}{{\mathrm{x}}^{\mathrm{n}}}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{n}\pi }{3}}+{\mathrm{e}}^{-\mathrm{i}\frac{2\mathrm{n}\pi }{3}}=2\cos \left(\frac{2\mathrm{n}\pi }{3}\right)$$$$\mathrm{take}\mathrm{n}=123456789$$$$\mathrm{if}\mathrm{x}={\mathrm{x}}_2\mathrm{we}\mathrm{get}{\mathrm{x}}^{\mathrm{n}}+\frac{1}{{\mathrm{x}}^{\mathrm{n}}}={\mathrm{e}}^{-\frac{2\mathrm{in}\pi }{3}}+{\mathrm{e}}^{\frac{2\mathrm{in}\pi }{3}}=2\cos \left(\frac{2\mathrm{n}\pi }{3}\right)$$$$\left(\mathrm{same}\mathrm{value}\right)$$

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