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Question Number 108480 by I want to learn more last updated on 17/Aug/20

Answered by Rasheed.Sindhi last updated on 17/Aug/20

$$x^2+x+1=0,x^{500}+\frac{1}{x^{500}}=?$$$$x^2+x+1=0\Rightarrow x(x+\frac{1}{x}+1)=0$$$$x\ne 0\Rightarrow x+\frac{1}{x}=-1$$$${(x+\frac{1}{x})}^2={(-1)}^2$$$$x^2+\frac{1}{x^2}=1-2=-1..............\left(i\right)$$$${(x+\frac{1}{x})}^3={(-1)}^3$$$$x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=-1$$$$x^3+\frac{1}{x^3}+3(-1)=-1$$$$x^3+\frac{1}{x^3}=2...................\left(ii\right)$$$$\left(i\right)\times \left(ii\right)$$$$x^5+\frac{1}{x^5}+x+\frac{1}{x}=-2$$$$x^5+\frac{1}{x^5}+(-1)=-2$$$$x^5+\frac{1}{x^5}=-1...................\left(iii\right)$$$${(x^5+\frac{1}{x^5})}^2={(-1)}^2$$$$x^{10}+\frac{1}{x^{10}}=-1$$$$x^n+\frac{1}{x^n}=-1$$$$\Rightarrow {(x^n+\frac{1}{x^n})}^2={(-1)}^2$$$$\Rightarrow x^{2n}+\frac{1}{x^{2n}}=1-2=-1$$$$\overline{⟨x^n+\frac{1}{x^n}=-1\Rightarrow x^{2n}+\frac{1}{x^{2n}}=-1⟩}$$$$Similarly$$$$\overline{⟨x^n+\frac{1}{x^n}=2\Rightarrow x^{3n}+\frac{1}{x^{3n}}=2⟩}$$$$$$$$....$$$$......$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}$$

Commented by udaythool last updated on 17/Aug/20

$$\mathrm{supper}$$

Answered by 1549442205PVT last updated on 17/Aug/20

$$\mathrm{From}\mathrm{the}\mathrm{hypothesis}\mathrm{we}\mathrm{get}$$$${\mathrm{x}}^2+\mathrm{x}+1=0\mathrm{\Delta }=1-4=-3={3\mathrm{i}}^2$$$$\Rightarrow \mathrm{x}=\frac{-1\pm \mathrm{i}\sqrt{3}}{2}=\cos \frac{2\pi }{3}\pm \mathrm{isin}\frac{2\pi }{3}$$$$\Rightarrow {\mathrm{x}}^{500}={(\cos \frac{2\pi }{3}\pm \mathrm{isin}\frac{2\pi }{3})}^{500}$$$$=\cos \frac{1000\pi }{3}\pm \sin \frac{1000\pi }{3}=\cos (333\pi +\frac{\pi }{3})\pm \mathrm{isin}(333\pi +\frac{\pi }{3})$$$$=-\cos \frac{\pi }{3}\mp \mathrm{isin}\frac{\pi }{3}=-\frac{1}{2}\mp \mathrm{i}\frac{\sqrt{3}}{2}$$$$=\frac{-1\mp \mathrm{i}\sqrt{3}}{2}\left(1\right)$$$$\Rightarrow \frac{1}{{\mathrm{x}}^{500}}=\frac{2}{-1\mp \mathrm{i}\sqrt{3}}=\frac{2(-1\pm \mathrm{i}\sqrt{3})}{{(-1)}^2-{\left(\mathrm{i}\sqrt{3}\right)}^2}=\frac{2(-1\pm \mathrm{i}\sqrt{3})}{1-{3\mathrm{i}}^2}$$$$=\frac{2(-1\pm \mathrm{i}\sqrt{3})}{4}=\frac{-1\pm \mathrm{i}\sqrt{3}}{2}\left(2\right)$$$$\mathrm{Adding}\mathrm{up}\mathrm{two}\mathrm{the}\mathrm{eqialities}\left(1\right)\left(2\right)$$$$\mathrm{we}\mathrm{get}{\mathrm{x}}^{500}+\frac{1}{{\mathrm{x}}^{500}}=\frac{-1\mp \mathrm{i}\sqrt{3}+-1\pm \mathrm{i}\sqrt{3}}{2}=-1$$$$\mathrm{Thus},{\mathbf{x}}^{500}+\frac{1}{{\mathbf{x}}^{500}}=\frac{-2}{2}=-1.$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{sir}$$

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

$$x^2+x+1=0$$$$x=e^{\frac{2i\pi }{3}}or{e}^{-\frac{2i\pi }{3}}$$$$x^{500}=e^{\frac{1000i\pi }{3}}=e^{-\frac{2\pi i}{3}}=-(\frac{1}{2}+\frac{\sqrt{3}}{2}i)$$$$\frac{1}{x^{500}}=e^{\frac{-1000\pi i}{3}}=e^{\frac{2i\pi }{3}}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$$$$x^{500}+\frac{1}{x^{500}}=-1$$$$Or$$$$x^{500}=e^{-\frac{1000i\pi }{3}}$$$$\frac{1}{x^{500}}=e^{\frac{1000i\pi }{3}}$$$$x^{500}+\frac{1}{x^{500}}=-1$$$$$$

Commented by udaythool last updated on 17/Aug/20

$$\mathrm{excellent}$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by mnjuly1970 last updated on 17/Aug/20

$$sol:x^2+x+1=0\Rightarrow x^3=1$$$$x^{500}+\frac{1}{x^{500}}={\left(x^3\right)}^{166}.x^2+{\left(\frac{1}{x^3}\right)}^{166}.\frac{1}{x^2}$$$$=x^2+\frac{1}{x^2}={(-1)}^2-2=-1\mathrm{♣}\mathrm{♣}\mathrm{♣}$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by Rasheed.Sindhi last updated on 17/Aug/20

$$x^2+x+1=0$$$$x=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm i\sqrt{3}}{2}$$$$x=\omega ,{\omega }^2$$$$x^{500}+\frac{1}{x^{500}}={\omega }^{500}+\frac{1}{{\omega }^{500}}$$$${\left({\omega }^3\right)}^{166}.{\omega }^2+\frac{1}{{\left({\omega }^3\right)}^{166}.{\omega }^2}$$$$={\omega }^2+\frac{1}{{\omega }^2}\times \frac{\omega }{\omega }={\omega }^2+\frac{\omega }{{\omega }^3}={\omega }^2+\omega =-1$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by floor(10²Eta[1]) last updated on 17/Aug/20

$${\mathrm{x}}^2=-(\mathrm{x}+1)$$$${\mathrm{x}}^4={(-(\mathrm{x}+1))}^2={\mathrm{x}}^2+2\mathrm{x}+1=-\mathrm{x}-1+2\mathrm{x}+1=\mathrm{x}$$$${\mathrm{x}}^5={\mathrm{x}}^4.\mathrm{x}=\mathrm{x}.\mathrm{x}={\mathrm{x}}^2=-\mathrm{x}-1$$$${\mathrm{x}}^{10}={(-(\mathrm{x}+1))}^2={\mathrm{x}}^2+2\mathrm{x}+1=-\mathrm{x}-1+2\mathrm{x}+1=\mathrm{x}$$$${\mathrm{x}}^{10}=\mathrm{x}\therefore {\mathrm{x}}^{500}={\mathrm{x}}^{50}={\left({\mathrm{x}}^{10}\right)}^5={\mathrm{x}}^5=-\mathrm{x}-1={\mathrm{x}}^2$$$$\Rightarrow {\mathrm{x}}^{500}+\frac{1}{{\mathrm{x}}^{500}}={\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}=\frac{{\mathrm{x}}^4+1}{{\mathrm{x}}^2}=\frac{\mathrm{x}+1}{-(\mathrm{x}+1)}=-1$$$$$$

Commented by Rasheed.Sindhi last updated on 17/Aug/20

$${\mathbf{e}}^{\mathbf{x}}\mathrm{cellent}\mathrm{approach}!$$

Commented by I want to learn more last updated on 17/Aug/20

$$\mathrm{Thanks}\mathrm{sir}$$

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