{ ((x^2 + x ((xy^2 ))^(1/(3 )) = 80 )),((y^2 + y ((x^2 y))^(1/(3 )) = 5 )) :} find x and y


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Question Number 95222 by bobhans last updated on 24/May/20
${ ((x^2 + x ((xy^2 ))^(1/(3 )) = 80 )),((y^2 + y ((x^2 y))^(1/(3 )) = 5 )) :} find x and y$
${\begin{cases} x^{2} + x \sqrt[3]{{x y}^{2}} = 80 \\ y^{2} + y \sqrt[3]{x^{2} y} = 5 \end{cases}$ $f i n d x a n d y$
	Answered by john santu last updated on 24/May/20
	$set (x)^(1/(3 )) = p & (y)^(1/(3 )) = q { ((p^6 + p^4 q^2 = 80)),((q^6 + p^2 q^4 = 5 )) :} { ((p^4 (p^2 +q^2 ) = 80)),((q^4 (p^2 +q^2 ) = 5 )) :} (p^4 /q^4 ) = 16 ⇒ p = ± 2q ⇒16q^4 (4q^2 +q^2 ) = 80 80q^4 = 80 ⇒q = ± 1 so we get p = ± 2 { (((x)^(1/(3 )) = ± 3 ⇒x = ± 8)),(((y)^(1/(3 )) = ± 1 ⇒ y = ± 1)) :} solution {(−8,−1),(8,1) }$
	$set \sqrt[3]{x} = p & \sqrt[3]{y} = q$ ${\begin{cases} p^{6} + p^{4} q^{2} = 80 \\ q^{6} + p^{2} q^{4} = 5 \end{cases}$ ${\begin{cases} p^{4} (p^{2} + q^{2}) = 80 \\ q^{4} (p^{2} + q^{2}) = 5 \end{cases}$ $\frac{p^{4}}{q^{4}} = 16 \Rightarrow p = \pm 2 q$ $\Rightarrow 16 q^{4} (4 q^{2} + q^{2}) = 80$ $80 q^{4} = 80 \Rightarrow q = \pm 1$ $so we get p = \pm 2$ ${\begin{cases} \sqrt[3]{x} = \pm 3 \Rightarrow x = \pm 8 \\ \sqrt[3]{y} = \pm 1 \Rightarrow y = \pm 1 \end{cases}$ $solution {(- 8, - 1), (8, 1)}$
	Commented by bobhans last updated on 24/May/20

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