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Question Number 179719 by cortano1 last updated on 01/Nov/22

Commented by Rasheed.Sindhi last updated on 01/Nov/22

$I s i t \sqrt[5]{x^{3} + 2 x} = \sqrt[3]{x^{5} - 4 x} ?$
Commented by mr W last updated on 01/Nov/22

$x = 0, x^{2} = 2 \Rightarrow x = \pm \sqrt{2}$
Commented by cortano1 last updated on 01/Nov/22

$yes the answer 0, \pm \sqrt{2}$
	Answered by Rasheed.Sindhi last updated on 01/Nov/22
	$((x^3 +2x))^(1/5) =((x^5 −2x))^(1/3) (((x^3 +2x))^(1/5) )^(15) =(((x^5 −2x))^(1/3) )^(15) (x^3 +2x )^3 =(x^5 −2x )^5 x^3 (x^2 +2)^3 =x^5 (x^4 −2)^5 x^3 (x^2 +2)^3 −x^5 (x^4 −2)^5 =0 x^3 {(x^2 +2)^3 −x^2 (x^4 −2)^5 }=0 x=0 ∣ (x^2 +2)^3 −x^2 (x^4 −2)^5 =0 x^3 (x+(2/x))^3 −x^2 ∙x^(10) (x^2 −(2/x^2 ))^5 =0 x^3 (x+(2/x))^3 −x^(12) (x^2 −(2/x^2 ))^5 =0 (x+(2/x))^3 −x^9 (x^2 −(2/x^2 ))^5 =0 [x≠0] Continue...$
	$\sqrt[5]{x^{3} + 2 x} = \sqrt[3]{x^{5} - 2 x}$ ${(\sqrt[5]{x^{3} + 2 x})}^{15} = {(\sqrt[3]{x^{5} - 2 x})}^{15}$ ${(x^{3} + 2 x)}^{3} = {(x^{5} - 2 x)}^{5}$ $x^{3} {(x^{2} + 2)}^{3} = x^{5} {(x^{4} - 2)}^{5}$ $x^{3} {(x^{2} + 2)}^{3} - x^{5} {(x^{4} - 2)}^{5} = 0$ $x^{3} {{(x^{2} + 2)}^{3} - x^{2} {(x^{4} - 2)}^{5}} = 0$ $x = 0 ∣ {(x^{2} + 2)}^{3} - x^{2} {(x^{4} - 2)}^{5} = 0$ $x^{3} {(x + \frac{2}{x})}^{3} - x^{2} \cdot x^{10} {(x^{2} - \frac{2}{x^{2}})}^{5} = 0$ $x^{3} {(x + \frac{2}{x})}^{3} - x^{12} {(x^{2} - \frac{2}{x^{2}})}^{5} = 0$ ${(x + \frac{2}{x})}^{3} - x^{9} {(x^{2} - \frac{2}{x^{2}})}^{5} = 0 [x \neq 0]$ $C o n t i n u e . . .$
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