Question Number 196154 by sonukgindia last updated on 19/Aug/23
Answered by mr W last updated on 19/Aug/23
$${x}^{\mathrm{2}} =−\mathrm{2}\left({x}+\mathrm{2}\right) \\ $$$${x}^{\mathrm{4}} =\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}\right)=\mathrm{4}\left(\mathrm{2}{x}\right)=\mathrm{8}{x} \\ $$$${x}^{\mathrm{6}} =\mathrm{8}{x}\left(−\mathrm{2}\right)\left({x}+\mathrm{2}\right)=−\mathrm{16}\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)=−\mathrm{16}×\left(−\mathrm{4}\right)=\mathrm{64}\:\checkmark \\ $$
Answered by BaliramKumar last updated on 19/Aug/23
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\:=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\right)\:=\mathrm{0}\left({x}−\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\:\neq\:\mathrm{2} \\ $$$${x}^{\mathrm{3}} \:−\:\mathrm{2}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$${x}^{\mathrm{3}} \:=\:\mathrm{8} \\ $$$$\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \:=\:\mathrm{8}^{\mathrm{2}} \\ $$$${x}^{\mathrm{6}} \:=\:\mathrm{64} \\ $$$$ \\ $$