Find all solutions of (x, y) such that x^3 − 3xy^2 = 2010 y^3 − 3x^2 y = 2009 x, y ∈ R


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Question Number 84909 by naka3546 last updated on 17/Mar/20

$F i n d a l l s o l u t i o n s o f (x, y) s u c h t h a t$ $x^{3} - 3 {x y}^{2} = 2010$ $y^{3} - 3 x^{2} y = 2009$ $x, y \in R$
Commented by john santu last updated on 17/Mar/20

$x^{3} - {3 xy}^{2} = 2010 [\div y^{3}]$ ${(\frac{x}{y})}^{3} - 3 (\frac{x}{y}) = \frac{2010}{y^{3}}$ $y^{3} - {3 x}^{2} y = 2009 [\div y^{3}]$ $1 - 3 {(\frac{x}{y})}^{2} = \frac{2009}{y^{3}}$ $let \frac{x}{y} = u \Rightarrow u^{3} - 3 u = \frac{2010}{y^{3}}$ $1 - {3 u}^{2} = \frac{2009}{y^{3}}$
Commented by john santu last updated on 17/Mar/20

$\Rightarrow \frac{1}{y^{3}} = \frac{1}{y^{3}}, \frac{u^{3} - 3 u}{2010} = \frac{1 - {3 u}^{2}}{2009}$ ${2009 u}^{3} - 6027 u = 2010 - {6030 u}^{2}$ ${2009 u}^{3} + {6030 u}^{2} - 6027 u - 2010 = 0$
Commented by MJS last updated on 17/Mar/20

$no ‘ ‘ {nice}^{″} solution$
Commented by john santu last updated on 17/Mar/20

$what the^{'} {nice}^{'} solution ?$
Commented by MJS last updated on 17/Mar/20

$we can only approximate$ $no solution of the shape a + \sqrt{b} or similar$
	Answered by MJS last updated on 17/Mar/20

	$x^{3} - 3 {x y}^{2} = 2010$ $y^{3} - 3 x^{2} y = 2009$ $let y = p x$ $(1 - 3 p^{2}) x^{3} = 2010$ $(p^{3} - 3 p) x^{3} = 2009$ $\Rightarrow$ $\frac{2010}{1 - 3 p^{2}} = \frac{2009}{p^{3} - 3 p}$ $\Rightarrow$ $p^{3} + \frac{2009}{670} p^{2} - 3 p - \frac{2009}{2010} = 0$ $p_{1} \approx - 3.73081$ $p_{2} \approx - . 267860$ $p_{3} \approx 1.00017$ $\Rightarrow$ $x_{1} \approx - 3.66718 \land y_{1} \approx 13.6815$ $x_{2} \approx 13.6832 \land y_{2} \approx - 3.66491$ $x_{3} \approx - 10.0150 \land y_{3} \approx - 10.0166$
	Answered by mind is power last updated on 19/Mar/20
	$−ix^3 +3ixy^2 =−2010i y^3 +3(ix)^2 y=2009 ⇒(y+ix)^3 =2009−2010i (y+ix)^3 =(√(2009^2 +2010^2 ))e^(−iarctan(((2010)/(2009)))+2ikπ) =re^(iθ_k ) y=r^(1/3) cos((θ_k /3)) x=r^(1/3) sin((θ_k /3)) reel solution y_k =((√(2009^2 +2010^2 )))^(1/3) .cos((1/3)arctan(((2010)/(2009)))+((2kπ)/3)) x_k =((√(2009^2 +2010^2 )))^(1/3) sin(−(1/3)arctan(((2010)/(2009)))+((2kπ)/3)),k∈{0,1,2}$
	$- {i x}^{3} + 3 {i x y}^{2} = - 2010 i$ $y^{3} + 3 {(i x)}^{2} y = 2009$ $\Rightarrow {(y + i x)}^{3} = 2009 - 2010 i$ ${(y + i x)}^{3} = \sqrt{2009^{2} + 2010^{2}} e^{- i a r c t a n (\frac{2010}{2009}) + 2 i k π} = {r e}^{i θ_{k}}$ $y = r^{\frac{1}{3}} c o s (\frac{θ_{k}}{3})$ $x = r^{\frac{1}{3}} s i n (\frac{θ_{k}}{3}) r e e l s o l u t i o n$ $y_{k} = {(\sqrt{2009^{2} + 2010^{2}})}^{\frac{1}{3}} . c o s (\frac{1}{3} a r c t a n (\frac{2010}{2009}) + \frac{2 k π}{3})$ $x_{k} = {(\sqrt{2009^{2} + 2010^{2}})}^{\frac{1}{3}} s i n (- \frac{1}{3} a r c t a n (\frac{2010}{2009}) + \frac{2 k π}{3}), k \in {0, 1, 2}$
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