solve^3 (√(1+x))+^3 (√(1−x))=2


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Question Number 64744 by mathmax by abdo last updated on 21/Jul/19

$s o l v e^{3} \sqrt{1 + x} +^{3} \sqrt{1 - x} = 2$
	Answered by MJS last updated on 21/Jul/19

	$a^{\frac{1}{3}} + b^{\frac{1}{3}} = c$ $a + 3 a^{\frac{2}{3}} b^{\frac{1}{3}} + 3 a^{\frac{1}{3}} b^{\frac{2}{3}} + b = c^{3}$ $a + 3 a^{\frac{1}{3}} b^{\frac{1}{3}} (a^{\frac{1}{3}} + b^{\frac{1}{3}}) + b = c^{3}$ $a + 3 a^{\frac{1}{3}} b^{\frac{1}{3}} c + b = c^{3}$ $3 a^{\frac{1}{3}} b^{\frac{1}{3}} c = c^{3} - a - b$ $27 {a b c}^{3} = {(c^{3} - a - b)}^{3}$ $216 (1 + x) (1 - x) = {(8 - (1 + x) - (1 - x))}^{3}$ $216 (1 - x^{2}) = 216$ $1 - x^{2} = 1$ $x = 0$
	Answered by behi83417@gmail.com last updated on 21/Jul/19
	$1+x=t^3 ,1−x=s^3 ⇒ { ((s+t=2)),((t^3 +s^3 =2)) :} ⇒(s+t)[(s+t)^2 −3st]=2 2(4−3st)=2⇒4−3st=1⇒st=1 z^2 −2z+1=0⇒z=1⇒ { ((1+x=1)),((1−x=1)) :}⇒x=0 .■$
	$1 + x = t^{3}, 1 - x = s^{3} \Rightarrow {\begin{cases} s + t = 2 \\ t^{3} + s^{3} = 2 \end{cases}$ $\Rightarrow (s + t) [{(s + t)}^{2} - 3 st] = 2$ $2 (4 - 3 st) = 2 \Rightarrow 4 - 3 st = 1 \Rightarrow st = 1$ $z^{2} - 2 z + 1 = 0 \Rightarrow z = 1 \Rightarrow {\begin{cases} 1 + x = 1 \\ 1 - x = 1 \end{cases} \Rightarrow x = 0 . ◼$
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