solve-3-1-x-3-1-x-2-

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} \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 .