Solve for x ((x^3 +3x−3))^(1/3) +((−x^3 −3x+5))^(1/3) =2 (An alteration of Q#202500)


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Question Number 202535 by Rasheed.Sindhi last updated on 29/Dec/23

$Solve for x$ $\sqrt[3]{x^{3} + 3 x - 3} + \sqrt[3]{- x^{3} - 3 x + 5} = 2$ $You can't use 'macro parameter character #' in math mode$
	Answered by mr W last updated on 29/Dec/23

	$\sqrt[n]{x^{3} + 3 x - 3} + \sqrt[n]{- x^{3} - 3 x + 5} = 2$ $n \in N^{⩾ 2}, x \in R$ $l e t x^{3} + 3 x - 4 = u$ $\sqrt[n]{1 + u} + \sqrt[n]{1 - u} = 2$ $u = 0 i s a t r i v i a l r o o t$ $\sqrt[n]{1 + u} i s s t r i c t l y i n c r e a s i n g .$ $\sqrt[n]{1 - u} i s s t r i c t l y d e c r e a s i n g .$ $\Rightarrow u = 0 i s o n e a n d t h e o n l y o n e r o o t$ $\Rightarrow x^{3} + 3 x - 4 = u = 0$ $\Rightarrow (x - 1) (x^{2} + x + 4) = 0$ $x^{2} + x + 4 = {(x + \frac{1}{2})}^{2} + \frac{15}{4} \neq 0$ $\Rightarrow x - 1 = 0 \Rightarrow x = 1 ✓$ $o r$ $\sqrt[n]{1 + u} + \sqrt[n]{1 - u} ⩽ 2$ $e q u a l i t y h o l d s w h e n \sqrt[n]{1 + u} = \sqrt[n]{1 - u},$ $i . e . 1 + u = 1 - u, \Rightarrow u = 0$
	Commented by mr W last updated on 29/Dec/23

	$i n c a s e n = 3 :$ $\sqrt[3]{1 + u} + \sqrt[3]{1 - u} = a w i t h 0 < a ⩽ 2$ $1 + u + 1 - u + 3 \sqrt[3]{(1 + u) (1 - u)} (\sqrt[3]{1 + u} + \sqrt[3]{1 - u}) = 2^{3}$ $2 + 3 a \sqrt[3]{(1 + u) (1 - u)} = a^{3}$ $\sqrt[3]{1 - u^{2}} = \frac{a^{3} - 2}{3 a}$ $\Rightarrow 1 - u^{2} = \frac{{(a^{3} - 2)}^{3}}{27 a^{3}}$ $\Rightarrow u = \pm \sqrt{1 - \frac{{(a^{3} - 2)}^{3}}{27 a^{3}}}$ $. . .$
	Commented by Rasheed.Sindhi last updated on 29/Dec/23

	$T han k s sir mr W!$
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