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Question Number 186636 by Noorzai last updated on 07/Feb/23

x^2 +x+1=0     x^(92) =?

$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{92}} =? \\ $$

Commented by Rasheed.Sindhi last updated on 07/Feb/23

x^(92) =ω,ω^2  where ω is complex cuberoot  of unity.

$${x}^{\mathrm{92}} =\omega,\omega^{\mathrm{2}} \:{where}\:\omega\:{is}\:{complex}\:{cuberoot} \\ $$$${of}\:{unity}. \\ $$

Commented by Noorzai last updated on 07/Feb/23

Giveanexplanation

$${Giveanexplanation} \\ $$

Answered by mr W last updated on 07/Feb/23

(1/x^2 )+(1/x)+1=0  ⇒(1/x)=((−1±(√3)i)/2)  (x−1)(x^2 +x+1)=0  x^3 −1=0  x^3 =1  x^(92) =(x^(93) /x)=(((x^3 )^(31) )/x)=(1/x)=((−1±(√3)i)/2)

$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{92}} =\frac{{x}^{\mathrm{93}} }{{x}}=\frac{\left({x}^{\mathrm{3}} \right)^{\mathrm{31}} }{{x}}=\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}} \\ $$

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