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

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

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

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

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

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

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