Menu Close

Find-all-solutions-of-x-y-such-that-x-3-3xy-2-2010-y-3-3x-2-y-2009-x-y-R-

Tinku Tara 0 Comments
Pin




Question Number 84909 by naka3546 last updated on 17/Mar/20
Find all solutions of (x, y) such that x^3 − 3xy^2 = 2010 y^3 − 3x^2 y = 2009 x, y ∈ R
$${Find}\:\:\:{all}\:\:{solutions}\:\:{of}\:\:\left({x},\:{y}\right)\:\:{such}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} \:−\:\mathrm{3}{xy}^{\mathrm{2}} \:\:=\:\:\mathrm{2010} \\ $$$$\:\:\:\:\:\:\:\:{y}^{\mathrm{3}} \:−\:\mathrm{3}{x}^{\mathrm{2}} {y}\:\:=\:\:\mathrm{2009} \\ $$$${x},\:{y}\:\:\in\:\:\mathbb{R} \\ $$
Commented by john santu last updated on 17/Mar/20
x^3 −3xy^2 = 2010 [÷y^3 ] ((x/y))^3 −3((x/y)) = ((2010)/y^3 ) y^3 −3x^2 y = 2009 [ ÷y^3 ] 1−3((x/y))^2 = ((2009)/y^3 ) let (x/y) = u ⇒ u^3 −3u = ((2010)/y^3 ) 1−3u^2 = ((2009)/y^3 )
$$\mathrm{x}^{\mathrm{3}} −\mathrm{3xy}^{\mathrm{2}} \:=\:\mathrm{2010}\:\left[\boldsymbol{\div}\mathrm{y}^{\mathrm{3}} \:\right]\: \\ $$$$\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{3}} −\mathrm{3}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\:=\:\frac{\mathrm{2010}}{\mathrm{y}^{\mathrm{3}} } \\ $$$$\mathrm{y}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} \mathrm{y}\:=\:\mathrm{2009}\:\left[\:\boldsymbol{\div}\mathrm{y}^{\mathrm{3}} \:\right] \\ $$$$\mathrm{1}−\mathrm{3}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{2009}}{\mathrm{y}^{\mathrm{3}} } \\ $$$$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{y}}\:=\:\mathrm{u}\:\Rightarrow\:\mathrm{u}^{\mathrm{3}} −\mathrm{3u}\:=\:\frac{\mathrm{2010}}{\mathrm{y}^{\mathrm{3}} } \\ $$$$\mathrm{1}−\mathrm{3u}^{\mathrm{2}} \:=\:\frac{\mathrm{2009}}{\mathrm{y}^{\mathrm{3}} }\: \\ $$
Commented by john santu last updated on 17/Mar/20
⇒ (1/y^3 ) = (1/y^3 ) , ((u^3 −3u)/(2010)) = ((1−3u^2 )/(2009)) 2009u^3 −6027u= 2010−6030u^2 2009u^3 +6030u^2 −6027u−2010 = 0
$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{3}} }\:,\:\frac{\mathrm{u}^{\mathrm{3}} −\mathrm{3u}}{\mathrm{2010}}\:=\:\frac{\mathrm{1}−\mathrm{3u}^{\mathrm{2}} }{\mathrm{2009}} \\ $$$$\mathrm{2009u}^{\mathrm{3}} −\mathrm{6027u}=\:\mathrm{2010}−\mathrm{6030u}^{\mathrm{2}} \\ $$$$\mathrm{2009u}^{\mathrm{3}} +\mathrm{6030u}^{\mathrm{2}} −\mathrm{6027u}−\mathrm{2010}\:=\:\mathrm{0} \\ $$
Commented by MJS last updated on 17/Mar/20
no “nice” solution
$$\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$
Commented by john santu last updated on 17/Mar/20
what the ′nice′ solution?
$$\mathrm{what}\:\mathrm{the}\:'\mathrm{nice}'\:\mathrm{solution}? \\ $$
Commented by MJS last updated on 17/Mar/20
we can only approximate no solution of the shape a+(√b) or similar
$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\mathrm{no}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shape}\:{a}+\sqrt{{b}}\:\mathrm{or}\:\mathrm{similar} \\ $$
Answered by MJS last updated on 17/Mar/20
x^3 −3xy^2 =2010 y^3 −3x^2 y=2009 let y=px (1−3p^2 )x^3 =2010 (p^3 −3p)x^3 =2009 ⇒ ((2010)/(1−3p^2 ))=((2009)/(p^3 −3p)) ⇒ p^3 +((2009)/(670))p^2 −3p−((2009)/(2010))=0 p_1 ≈−3.73081 p_2 ≈−.267860 p_3 ≈1.00017 ⇒ x_1 ≈−3.66718∧y_1 ≈13.6815 x_2 ≈13.6832∧y_2 ≈−3.66491 x_3 ≈−10.0150∧y_3 ≈−10.0166
$${x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{2010} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}=\mathrm{2009} \\ $$$$\mathrm{let}\:{y}={px} \\ $$$$\left(\mathrm{1}−\mathrm{3}{p}^{\mathrm{2}} \right){x}^{\mathrm{3}} =\mathrm{2010} \\ $$$$\left({p}^{\mathrm{3}} −\mathrm{3}{p}\right){x}^{\mathrm{3}} =\mathrm{2009} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{2010}}{\mathrm{1}−\mathrm{3}{p}^{\mathrm{2}} }=\frac{\mathrm{2009}}{{p}^{\mathrm{3}} −\mathrm{3}{p}} \\ $$$$\Rightarrow \\ $$$${p}^{\mathrm{3}} +\frac{\mathrm{2009}}{\mathrm{670}}{p}^{\mathrm{2}} −\mathrm{3}{p}−\frac{\mathrm{2009}}{\mathrm{2010}}=\mathrm{0} \\ $$$${p}_{\mathrm{1}} \approx−\mathrm{3}.\mathrm{73081} \\ $$$${p}_{\mathrm{2}} \approx−.\mathrm{267860} \\ $$$${p}_{\mathrm{3}} \approx\mathrm{1}.\mathrm{00017} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{3}.\mathrm{66718}\wedge{y}_{\mathrm{1}} \approx\mathrm{13}.\mathrm{6815} \\ $$$${x}_{\mathrm{2}} \approx\mathrm{13}.\mathrm{6832}\wedge{y}_{\mathrm{2}} \approx−\mathrm{3}.\mathrm{66491} \\ $$$${x}_{\mathrm{3}} \approx−\mathrm{10}.\mathrm{0150}\wedge{y}_{\mathrm{3}} \approx−\mathrm{10}.\mathrm{0166} \\ $$
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}
$$−{ix}^{\mathrm{3}} +\mathrm{3}{ixy}^{\mathrm{2}} =−\mathrm{2010}{i} \\ $$$${y}^{\mathrm{3}} +\mathrm{3}\left({ix}\right)^{\mathrm{2}} {y}=\mathrm{2009} \\ $$$$\Rightarrow\left({y}+{ix}\right)^{\mathrm{3}} =\mathrm{2009}−\mathrm{2010}{i} \\ $$$$\left({y}+{ix}\right)^{\mathrm{3}} =\sqrt{\mathrm{2009}^{\mathrm{2}} +\mathrm{2010}^{\mathrm{2}} }{e}^{−{iarctan}\left(\frac{\mathrm{2010}}{\mathrm{2009}}\right)+\mathrm{2}{ik}\pi} ={re}^{{i}\theta_{{k}} } \\ $$$${y}={r}^{\frac{\mathrm{1}}{\mathrm{3}}} {cos}\left(\frac{\theta_{{k}} }{\mathrm{3}}\right) \\ $$$${x}={r}^{\frac{\mathrm{1}}{\mathrm{3}}} {sin}\left(\frac{\theta_{{k}} }{\mathrm{3}}\right)\:\:\:{reel}\:{solution} \\ $$$${y}_{{k}} =\left(\sqrt{\mathrm{2009}^{\mathrm{2}} +\mathrm{2010}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} .{cos}\left(\frac{\mathrm{1}}{\mathrm{3}}{arctan}\left(\frac{\mathrm{2010}}{\mathrm{2009}}\right)+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$${x}_{{k}} =\left(\sqrt{\mathrm{2009}^{\mathrm{2}} +\mathrm{2010}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {sin}\left(−\frac{\mathrm{1}}{\mathrm{3}}{arctan}\left(\frac{\mathrm{2010}}{\mathrm{2009}}\right)+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right),{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2}\right\} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *