Question Number 186636 by Noorzai last updated on 07/Feb/23
$${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{x}^{\mathrm{92}} =? \\ $$
Commented by Rasheed.Sindhi last updated on 07/Feb/23
$${x}^{\mathrm{92}} =\omega,\omega^{\mathrm{2}} \:{where}\:\omega\:{is}\:{complex}\:{cuberoot} \\ $$$${of}\:{unity}. \\ $$
Commented by Noorzai last updated on 07/Feb/23
$${Giveanexplanation} \\ $$
Answered by mr W last updated on 07/Feb/23
$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} =\mathrm{1} \\ $$$${x}^{\mathrm{92}} =\frac{{x}^{\mathrm{93}} }{{x}}=\frac{\left({x}^{\mathrm{3}} \right)^{\mathrm{31}} }{{x}}=\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{3}}{i}}{\mathrm{2}} \\ $$