(((x+4)^2 ))^(1/(3 )) + 4 (((x−3)^2 ))^(1/(3 )) + 5 ((x^2 +x−12))^(1/(3 )) = 0


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Question Number 96131 by bemath last updated on 30/May/20

$\sqrt[3]{{(x + 4)}^{2}} + 4 \sqrt[3]{{(x - 3)}^{2}} + 5 \sqrt[3]{x^{2} + x - 12} = 0$
	Answered by john santu last updated on 30/May/20
	$let ((x+4))^(1/(3 )) = u & ((x−3))^(1/(3 )) = v ⇒u^2 +4v^2 +5uv = 0 (u+4v)(u+v) = 0 { ((u = −v)),((u=−4v)) :} (1) ((x+4))^(1/(3 )) = −((x−3))^(1/(3 )) x+4 = −x+3 ⇒ x=−(1/2) (2) ((x+4))^(1/(3 )) = −4 ((x−3))^(1/(3 )) x+4 = −64x+192 65x = 184 ; x = ((184)/(65)) .$
	$let \sqrt[3]{x + 4} = u & \sqrt[3]{x - 3} = v$ $\Rightarrow u^{2} + 4 v^{2} + 5 u v = 0$ $(u + 4 v) (u + v) = 0$ ${\begin{cases} u = - v \\ u = - 4 v \end{cases}$ $(1) \sqrt[3]{x + 4} = - \sqrt[3]{x - 3}$ $x + 4 = - x + 3 \Rightarrow x = - \frac{1}{2}$ $(2) \sqrt[3]{x + 4} = - 4 \sqrt[3]{x - 3}$ $x + 4 = - 64 x + 192$ $65 x = 184; x = \frac{184}{65} .$
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