Question Number 213636 by hardmath last updated on 11/Nov/24
$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}^{\mathrm{10}} \:=\:? \\ $$
Answered by A5T last updated on 11/Nov/24
$${x}^{\mathrm{2}} ={x}−\mathrm{1}\Rightarrow{x}^{\mathrm{3}} ={x}^{\mathrm{2}} −{x}=−\mathrm{1}\Rightarrow{x}^{\mathrm{9}} =−\mathrm{1}\Rightarrow{x}^{\mathrm{10}} =−{x} \\ $$$${x}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}\Rightarrow−{x}=\frac{−\mathrm{1}}{\mathrm{2}}\overset{−} {+}\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Answered by Frix last updated on 12/Nov/24
$${x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{Borrow}\:{x}=−\mathrm{1} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$${x}=−\mathrm{1}\:\left(\mathrm{Return}\:\mathrm{with}\:\mathrm{thanks}\right) \\ $$$${x}=\mathrm{e}^{\pm\mathrm{i}\frac{\pi}{\mathrm{3}}} \\ $$$$\Rightarrow\:{x}^{\mathrm{10}} =\mathrm{e}^{\pm\frac{\mathrm{10}\pi}{\mathrm{3}}} =\mathrm{e}^{\mp\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} =−{x} \\ $$