x^2 =(2)^(1/5) +y y^2 =(2)^(1/5) +x x∙y=? x≠y


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Question Number 189446 by mathlove last updated on 16/Mar/23

$x^{2} = \sqrt[5]{2} + y$ $y^{2} = \sqrt[5]{2} + x$ $x \cdot y = ? x \neq y$
	Answered by cortano12 last updated on 16/Mar/23
	$⇒x^2 −y^2 = y−x ⇒(x−y)(x+y)+(x−y)=0 ⇒(x−y){x+y+1}=0 ⇒y=−1−x ; x^2 = (2)^(1/5) −1−x ⇒x^2 +x=(2)^(1/5) −1 ∴ xy = x(−1−x)=−x^2 −x=1−(2)^(1/5)$
	$\Rightarrow x^{2} - y^{2} = y - x$ $\Rightarrow (x - y) (x + y) + (x - y) = 0$ $\Rightarrow (x - y) {x + y + 1} = 0$ $\Rightarrow y = - 1 - x; x^{2} = \sqrt[5]{2} - 1 - x$ $\Rightarrow x^{2} + x = \sqrt[5]{2} - 1$ $∴ xy = x (- 1 - x) = - x^{2} - x = 1 - \sqrt[5]{2}$
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