Question Number 62281 by behi83417@gmail.com last updated on 19/Jun/19
![{ ((x^3 +y^3 =3xy)),((x^4 +y^4 =4xy)) :} [x,y≠0]](https://www.tinkutara.com/question/Q62281.png)
$$\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\mathrm{3}\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{y}}^{\mathrm{4}} =\mathrm{4}\boldsymbol{\mathrm{xy}}}\end{cases}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\neq\mathrm{0}\right] \\ $$
Answered by mr W last updated on 19/Jun/19
![let u=x+y, v=xy x^3 +y^3 =3xy (x+y)^3 =x^3 +y^3 +3xy(x+y) (x+y)^3 =3xy+3xy(x+y) u^3 =3v(1+u) ⇒v=(u^3 /(3(1+u))) x^4 +y^4 =(x^2 +y^2 )^2 −2x^2 y^2 =[(x+y)^2 −2xy]^2 −2(xy)^2 =4xy ⇒(u^2 −2v)^2 −2v^2 =4v ⇒[u^2 −((2u^3 )/(3(1+u)))]^2 −2×(u^6 /(9(1+u)^2 ))=4×(u^3 /(3(1+u))) ⇒[3u^2 (1+u)−2u^3 ]^2 −2u^6 =12u^3 (1+u) ⇒u^4 (3+u)^2 −2u^6 =12u^3 (1+u) ⇒u=0⇒x=y=0 ⇒u(3+u)^2 −2u^3 =12(1+u) ⇒u^3 −6u^2 +3u+12=0 let u=s+2 ⇒s^3 −9s+2=0 Δ=(−3)^3 +1^2 <0 ⇒3 real roots ⇒s=2(√3) sin ((1/3) sin^(−1) (1/(3(√3)))+((2kπ)/3)) ⇒u=2+2(√3) sin ((1/3) sin^(−1) (1/(3(√3)))+((2kπ)/3)) ⇒v=(u^3 /(3(1+u))) ⇒(x−y)^2 =(x+y)^2 −4xy=u^2 −4v ⇒x−y=±(√(u^2 −4v)) ⇒x+y=u ⇒x=((u±(√(u^2 −4v)))/2) ⇒y=((u∓(√(u^2 −4v)))/2) { ((x=1.426766 / 0.796696 / 2.44101+0.79718i / 2.44101−0.79718i / −0.55274+1.99092i /−0.55274−1.99092i )),((y=0.796696 / 1.426766 / 2.44101−0.79718i / 2.44101+0.79718i / −0.55274−1.99092i / −0.55274+1.99092i )) :}](https://www.tinkutara.com/question/Q62323.png)
$${let}\:{u}={x}+{y},\:{v}={xy} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{3}{xy} \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} =\mathrm{3}{xy}+\mathrm{3}{xy}\left({x}+{y}\right) \\ $$$${u}^{\mathrm{3}} =\mathrm{3}{v}\left(\mathrm{1}+{u}\right) \\ $$$$\Rightarrow{v}=\frac{{u}^{\mathrm{3}} }{\mathrm{3}\left(\mathrm{1}+{u}\right)} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} =\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\left[\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}\right]^{\mathrm{2}} −\mathrm{2}\left({xy}\right)^{\mathrm{2}} =\mathrm{4}{xy} \\ $$$$\Rightarrow\left({u}^{\mathrm{2}} −\mathrm{2}{v}\right)^{\mathrm{2}} −\mathrm{2}{v}^{\mathrm{2}} =\mathrm{4}{v} \\ $$$$\Rightarrow\left[{u}^{\mathrm{2}} −\frac{\mathrm{2}{u}^{\mathrm{3}} }{\mathrm{3}\left(\mathrm{1}+{u}\right)}\right]^{\mathrm{2}} −\mathrm{2}×\frac{{u}^{\mathrm{6}} }{\mathrm{9}\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }=\mathrm{4}×\frac{{u}^{\mathrm{3}} }{\mathrm{3}\left(\mathrm{1}+{u}\right)} \\ $$$$\Rightarrow\left[\mathrm{3}{u}^{\mathrm{2}} \left(\mathrm{1}+{u}\right)−\mathrm{2}{u}^{\mathrm{3}} \right]^{\mathrm{2}} −\mathrm{2}{u}^{\mathrm{6}} =\mathrm{12}{u}^{\mathrm{3}} \left(\mathrm{1}+{u}\right) \\ $$$$\Rightarrow{u}^{\mathrm{4}} \left(\mathrm{3}+{u}\right)^{\mathrm{2}} −\mathrm{2}{u}^{\mathrm{6}} =\mathrm{12}{u}^{\mathrm{3}} \left(\mathrm{1}+{u}\right) \\ $$$$\Rightarrow{u}=\mathrm{0}\Rightarrow{x}={y}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow{u}\left(\mathrm{3}+{u}\right)^{\mathrm{2}} −\mathrm{2}{u}^{\mathrm{3}} =\mathrm{12}\left(\mathrm{1}+{u}\right) \\ $$$$\Rightarrow{u}^{\mathrm{3}} −\mathrm{6}{u}^{\mathrm{2}} +\mathrm{3}{u}+\mathrm{12}=\mathrm{0} \\ $$$${let}\:{u}={s}+\mathrm{2} \\ $$$$\Rightarrow{s}^{\mathrm{3}} −\mathrm{9}{s}+\mathrm{2}=\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{3}\right)^{\mathrm{3}} +\mathrm{1}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\mathrm{3}\:{real}\:{roots} \\ $$$$\Rightarrow{s}=\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow{u}=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow{v}=\frac{{u}^{\mathrm{3}} }{\mathrm{3}\left(\mathrm{1}+{u}\right)} \\ $$$$\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}={u}^{\mathrm{2}} −\mathrm{4}{v} \\ $$$$\Rightarrow{x}−{y}=\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}} \\ $$$$\Rightarrow{x}+{y}={u} \\ $$$$\Rightarrow{x}=\frac{{u}\pm\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{{u}\mp\sqrt{{u}^{\mathrm{2}} −\mathrm{4}{v}}}{\mathrm{2}} \\ $$$$\begin{cases}{{x}=\mathrm{1}.\mathrm{426766}\:/\:\mathrm{0}.\mathrm{796696}\:/\:\mathrm{2}.\mathrm{44101}+\mathrm{0}.\mathrm{79718}{i}\:/\:\mathrm{2}.\mathrm{44101}−\mathrm{0}.\mathrm{79718}{i}\:/\:−\mathrm{0}.\mathrm{55274}+\mathrm{1}.\mathrm{99092}{i}\:/−\mathrm{0}.\mathrm{55274}−\mathrm{1}.\mathrm{99092}{i}\:}\\{{y}=\mathrm{0}.\mathrm{796696}\:/\:\mathrm{1}.\mathrm{426766}\:/\:\mathrm{2}.\mathrm{44101}−\mathrm{0}.\mathrm{79718}{i}\:/\:\mathrm{2}.\mathrm{44101}+\mathrm{0}.\mathrm{79718}{i}\:/\:−\mathrm{0}.\mathrm{55274}−\mathrm{1}.\mathrm{99092}{i}\:/\:−\mathrm{0}.\mathrm{55274}+\mathrm{1}.\mathrm{99092}{i}\:}\end{cases} \\ $$
Commented by behi83417@gmail.com last updated on 19/Jun/19
$$\mathrm{thanks}\:\mathrm{in}\:\mathrm{advance}\:\mathrm{dear}\:\mathrm{master}. \\ $$$$\mathrm{nice}\:\mathrm{and}\:\mathrm{smart}\:\mathrm{solution}. \\ $$
Commented by behi83417@gmail.com last updated on 19/Jun/19
