xy + (xy^2 )^(1/3) = y^2 Solve


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Question Number 182534 by Mastermind last updated on 10/Dec/22

$xy + {({xy}^{2})}^{\frac{1}{3}} = y^{2}$ $Solve$
Commented by Frix last updated on 10/Dec/22

$x = y = 0$ $y = p^{3} x \land x = \frac{1}{p (p^{3} - 1)}$
Commented by mr W last updated on 11/Dec/22

$i t i s a c u r v e . c a n n o t b e s o l v e d!$
Commented by Frix last updated on 11/Dec/22

$Yes of vourse with p as parameter . I tbought$ $this was obvious .$
	Answered by manxsol last updated on 11/Dec/22
	$Right Frix, develop your approach and put the condition for p and solution set. Thanks Sir W for correction PROCESS: parameterized curve not solved if y=x⇏ x^2 +x=x^2 condition y≠x y=p^3 x x(p^3 x)+(xp^6 x^2 )^(1/3) =p^6 x^2 x^2 p^3 +xp^2 =p^6 x^2 x=0⇒y=0 (0;0) xp^3 +p^2 =p^6 x x=(p^2 /(p^6 −p^3 ))=(1/(p(p^3 −1))) parameterized f(p)={(0;0) ((1/(p(p^3 −1))); (p^2 /((p^3 −1)))) p≠1 p≠0$
	$R i g h t F r i x, d e v e l o p y o u r$ $a p p r o a c h a n d p u t t h e$ $c o n d i t i o n f o r p a n d$ $s o l u t i o n s e t .$ $T h a n k s S i r W$ $f o r c o r r e c t i o n$ $P R O C E S S : p a r a m e t e r i z e d$ $c u r v e$ $n o t s o l v e d$ $i f y = x ⇏ x^{2} + x = x^{2}$ $c o n d i t i o n y \neq x$ $y = p^{3} x$ $x (p^{3} x) + {({x p}^{6} x^{2})}^{\frac{1}{3}} = p^{6} x^{2}$ $x^{2} p^{3} + {x p}^{2} = p^{6} x^{2}$ $x = 0 \Rightarrow y = 0 (0; 0)$ ${x p}^{3} + p^{2} = p^{6} x$ $x = \frac{p^{2}}{p^{6} - p^{3}} = \frac{1}{p (p^{3} - 1)}$ $p a r a m e t e r i z e d$ $f (p) = {(0; 0)$ $(\frac{1}{p (p^{3} - 1)}; \frac{p^{2}}{(p^{3} - 1)})$ $p \neq 1 p \neq 0$
	Commented by Mastermind last updated on 11/Dec/22

	$Sorry, there was a mistake in the question$
	Commented by Mastermind last updated on 11/Dec/22

	$Check the next slide for new ques .$ $l^{'} m sorry$
	Commented by manxsol last updated on 11/Dec/22

	$h a s n o t b e e n r e s o l v e d,$ $h a s b e e n p a r a m e t e r i z e d$
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