Question Number 19595 by khamizan833@yahoo.com last updated on 13/Aug/17
$$\mathrm{For}\:{x}\:\in\:\mathrm{R},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{below}! \\ $$$$\left(\mathrm{2}^{{x}} \:−\:\mathrm{4}\right)^{\mathrm{3}} \:+\:\left(\mathrm{4}^{{x}} \:−\:\mathrm{2}\right)^{\mathrm{3}} \:=\:\left(\mathrm{4}^{{x}} \:+\:\mathrm{2}^{{x}} \:−\:\mathrm{6}\right)^{\mathrm{3}} \\ $$
Commented by khamizan833@yahoo.com last updated on 13/Aug/17
$$\mathrm{show}\:\mathrm{your}\:\mathrm{work},\mathrm{please}. \\ $$
Answered by Tinkutara last updated on 13/Aug/17
$$\mathrm{Let}\:\mathrm{2}^{{x}} \:=\:{y}. \\ $$$$\left({y}\:−\:\mathrm{4}\right)^{\mathrm{3}} \:+\:\left({y}^{\mathrm{2}} \:−\:\mathrm{2}\right)^{\mathrm{3}} \:+\:\left(\mathrm{6}\:−\:{y}\:−\:{y}^{\mathrm{2}} \right)^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{where}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{so} \\ $$$${a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:=\:\mathrm{3}{abc}\:=\:\mathrm{0}\:\mathrm{as}\:\mathrm{given}. \\ $$$$\left({y}\:−\:\mathrm{4}\right)\left({y}^{\mathrm{2}} \:−\:\mathrm{2}\right)\left(\mathrm{6}\:−\:{y}\:−\:{y}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{y}\:=\:\mathrm{4},\:\pm\:\sqrt{\mathrm{2}},\:\mathrm{2} \\ $$$$\mathrm{But}\:\mathrm{only}\:\mathrm{positive}\:\mathrm{values}\:\mathrm{of}\:{y}\:\mathrm{are} \\ $$$$\mathrm{allowed}.\:\mathrm{So}\:\mathrm{2}^{{x}} \:=\:\mathrm{4},\:\sqrt{\mathrm{2}},\:\mathrm{2} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{2},\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{1} \\ $$
Commented by khamizan833@yahoo.com last updated on 13/Aug/17
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\mathrm{god}\:\mathrm{bless}\:\mathrm{you}. \\ $$