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Question Number 148257 by liberty last updated on 26/Jul/21

	Answered by Olaf_Thorendsen last updated on 26/Jul/21
	$(2^x −4)^3 +(4^x −2)^3 = (4^x +2^x −6)^3 Let u = 2^x (u−4)^3 +(u^2 −2)^3 = (u^2 +u−6)^3 (u−4)^3 +(u^2 −2)^3 = [(u^2 −2)+(u−4)]^3 0 = 3(u^2 −2)^2 (u−4)+3(u^2 −2)(u−4)^2 (u^2 −2)(u−4)(u^2 −2+u−4) = 0 (u+(√2))(u−(√2))(u−4)(u+3)(u−2) = 0 u = ±(√2), 4,−3, 2 −(√2), −3 impossible because u = 2^x >0 u = (√2), 2, 4 ⇒ x = (1/2), 1, 2 S = {(1/2), 1, 2}$
	${(2^{x} - 4)}^{3} + {(4^{x} - 2)}^{3} = {(4^{x} + 2^{x} - 6)}^{3}$ $Let u = 2^{x}$ ${(u - 4)}^{3} + {(u^{2} - 2)}^{3} = {(u^{2} + u - 6)}^{3}$ ${(u - 4)}^{3} + {(u^{2} - 2)}^{3} = {[(u^{2} - 2) + (u - 4)]}^{3}$ $0 = 3 {(u^{2} - 2)}^{2} (u - 4) + 3 (u^{2} - 2) {(u - 4)}^{2}$ $(u^{2} - 2) (u - 4) (u^{2} - 2 + u - 4) = 0$ $(u + \sqrt{2}) (u - \sqrt{2}) (u - 4) (u + 3) (u - 2) = 0$ $u = \pm \sqrt{2}, 4, - 3, 2$ $- \sqrt{2}, - 3 impossible because u = 2^{x} > 0$ $u = \sqrt{2}, 2, 4 \Rightarrow x = \frac{1}{2}, 1, 2$ $S = {\frac{1}{2}, 1, 2}$
	Answered by Rasheed.Sindhi last updated on 26/Jul/21

	${(\underset{u}{\underset{⏟}{2^{x} - 4}})}^{3} + {(\underset{v}{\underset{⏟}{4^{x} - 2}})}^{3} = {(4^{x} + 2^{x} - 6)}^{3}$ $u^{3} + v^{3} = {(u + v)}^{3}$ $u^{3} + v^{3} = u^{3} + v^{3} + 2 u v (u + v)$ $u v (u + v) = 0$ $u = 0 ∣ v = 0 ∣ u + v = 0$ $2^{x} - 4 = 0 ∣ 4^{x} - 2 = 0 ∣ 2^{x} - 4 = - 4^{x} + 2$ $2^{x} = 2^{2} ∣ 2^{2 x} = 2^{1} ∣ \underset{y}{\underset{⏟}{2^{x}}} + 2^{2 x} - 6 = 0$ $x = 2 ∣ 2 x = 1 ∣ y^{2} + y - 6 = 0$ $∣ x = 1 / 2 ∣ y^{2} + y - 6 = 0$ $∣ (y + 3) (y - 2) = 0$ $∣ y = - 3 ∣ y = 2$ $∣ 2^{x} = - 3 ∣ 2^{x} = 2^{1}$ $∣ n o t r e a l ∣ x = 1$ $2^{x} = - 3$ $x \log (2) = \log (- 3)$ $x = \frac{\log (- 3)}{\log (2)} = 1.585 + 4.532 i$ $x = 1, 2, 1 / 2, 1.585 + 4.532 i$
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