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Question Number 19595 by khamizan833@yahoo.com last updated on 13/Aug/17

For x ∈ R, solve the equation below!  (2^x  − 4)^3  + (4^x  − 2)^3  = (4^x  + 2^x  − 6)^3

$$\mathrm{For}\:{x}\:\in\:\mathrm{R},\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{below}! \\ $$$$\left(\mathrm{2}^{{x}} \:−\:\mathrm{4}\right)^{\mathrm{3}} \:+\:\left(\mathrm{4}^{{x}} \:−\:\mathrm{2}\right)^{\mathrm{3}} \:=\:\left(\mathrm{4}^{{x}} \:+\:\mathrm{2}^{{x}} \:−\:\mathrm{6}\right)^{\mathrm{3}} \\ $$

Commented by khamizan833@yahoo.com last updated on 13/Aug/17

show your work,please.

$$\mathrm{show}\:\mathrm{your}\:\mathrm{work},\mathrm{please}. \\ $$

Answered by Tinkutara last updated on 13/Aug/17

Let 2^x  = y.  (y − 4)^3  + (y^2  − 2)^3  + (6 − y − y^2 )^3  = 0  It is of the form where a + b + c = 0 so  a^3  + b^3  + c^3  = 3abc = 0 as given.  (y − 4)(y^2  − 2)(6 − y − y^2 ) = 0  ⇒ y = 4, ± (√2), 2  But only positive values of y are  allowed. So 2^x  = 4, (√2), 2  ⇒ x = 2, (1/2), 1

$$\mathrm{Let}\:\mathrm{2}^{{x}} \:=\:{y}. \\ $$$$\left({y}\:−\:\mathrm{4}\right)^{\mathrm{3}} \:+\:\left({y}^{\mathrm{2}} \:−\:\mathrm{2}\right)^{\mathrm{3}} \:+\:\left(\mathrm{6}\:−\:{y}\:−\:{y}^{\mathrm{2}} \right)^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\mathrm{where}\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{so} \\ $$$${a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:=\:\mathrm{3}{abc}\:=\:\mathrm{0}\:\mathrm{as}\:\mathrm{given}. \\ $$$$\left({y}\:−\:\mathrm{4}\right)\left({y}^{\mathrm{2}} \:−\:\mathrm{2}\right)\left(\mathrm{6}\:−\:{y}\:−\:{y}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\Rightarrow\:{y}\:=\:\mathrm{4},\:\pm\:\sqrt{\mathrm{2}},\:\mathrm{2} \\ $$$$\mathrm{But}\:\mathrm{only}\:\mathrm{positive}\:\mathrm{values}\:\mathrm{of}\:{y}\:\mathrm{are} \\ $$$$\mathrm{allowed}.\:\mathrm{So}\:\mathrm{2}^{{x}} \:=\:\mathrm{4},\:\sqrt{\mathrm{2}},\:\mathrm{2} \\ $$$$\Rightarrow\:{x}\:=\:\mathrm{2},\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{1} \\ $$

Commented by khamizan833@yahoo.com last updated on 13/Aug/17

thank you sir.god bless you.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\mathrm{god}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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