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Question Number 168131 by daus last updated on 04/Apr/22

Commented by daus last updated on 04/Apr/22

$h e l p m e t o s o l v e a b o v e e q u a t i o n$
	Answered by benhamimed last updated on 04/Apr/22
	$(d)⇔ { ((x^3 =t)),((t^2 −2t−8=0)) :} t=−2 ;t=4 S={−(2)^(1/3) ;(4)^(1/3) } (e)→x^6 −6x^3 −16=0 ⇔ { ((x^3 =t)),((t^2 −6t−16=0)) :} t=−2;8 S={−(2)^(1/3) ;2}$
	$(d) \Leftrightarrow {\begin{cases} x^{3} = t \\ t^{2} - 2 t - 8 = 0 \end{cases}$ $t = - 2; t = 4$ $S = {- \sqrt[3]{2}; \sqrt[3]{4}}$ $(e) \to x^{6} - 6 x^{3} - 16 = 0$ $\Leftrightarrow {\begin{cases} x^{3} = t \\ t^{2} - 6 t - 16 = 0 \end{cases}$ $t = - 2; 8$ $S = {- \sqrt[3]{2}; 2}$
	Answered by MJS_new last updated on 04/Apr/22

	$x^{6} - 2 x^{3} - 8 = 0$ $(x^{3} - 4) (x^{3} + 2) = 0$ $x_{1} = \sqrt[3]{4}$ $x_{2, 3} = - \frac{1 \pm \sqrt{3} i}{\sqrt[3]{2}}$ $x_{4} = - \sqrt[3]{2}$ $x_{5, 6} = \frac{1 \pm \sqrt{3} i}{\sqrt[3]{4}}$ $x^{2} = \frac{2 (3 x^{3} + 8)}{x^{4}}$ $x^{6} - 6 x^{3} - 16 = 0$ $(x - 2) (x^{2} + 2 x + 4) (x^{3} + 2) = 0$ $x_{1} = 2$ $x_{2, 3} = - 1 \pm \sqrt{3} i$ $x_{4} = - \sqrt[3]{2}$ $x_{5, 6} = \frac{1 \pm \sqrt{3} i}{\sqrt[3]{4}}$
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