Question Number 44876 by Tinkutara last updated on 05/Oct/18
Answered by ajfour last updated on 06/Oct/18
![let z_1 = x , z_2 = y (just calling) { (( x(x^2 −3y^2 )=2)),(( y(3x^2 −y^2 )=1)) :} Then x^3 −3xy^2 = 6x^2 y−2y^3 let y=tx 1−3t^2 = 6t−2t^3 or 2t^3 −3t^2 −6t+1 = 0 (from here let t = t_1 , t_2 ,t_3 ) Now x^2 −3y^2 = (2/x) 3x^2 −y^2 = (1/y) ⇒ 2x^2 +2y^2 = (1/y)+(2/x) = 2q since y=tx ⇒ x^2 (1+t^2 )=(1/x)((1/t)+1) = 2q ⇒ x^3 = ((t+1)/(t^3 +t)) ⇒ 2q = (1+t^2 )(((t+1)/(t^3 +t)))^(2/3) q=(x^2 +y^2 )= (1/2)[ (((1+t^2 )(1+t)^2 )/t^2 ) ]^(1/3) .](https://www.tinkutara.com/question/Q44888.png)
$${let}\:\:{z}_{\mathrm{1}} =\:{x}\:\:,\:\:{z}_{\mathrm{2}} \:=\:{y}\:\:\:\:\left({just}\:{calling}\right) \\ $$$$\:\:\:\:\begin{cases}{\:\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{y}}^{\mathrm{2}} \right)=\mathrm{2}}\\{\:\boldsymbol{{y}}\left(\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{y}}^{\mathrm{2}} \right)=\mathrm{1}}\end{cases} \\ $$$${Then}\:\:\:\:{x}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} \:=\:\mathrm{6}{x}^{\mathrm{2}} {y}−\mathrm{2}{y}^{\mathrm{3}} \\ $$$${let}\:\:\:{y}={tx} \\ $$$$\:\:\:\:\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} \:=\:\mathrm{6}{t}−\mathrm{2}{t}^{\mathrm{3}} \\ $$$${or}\:\:\:\:\:\:\:\mathrm{2}{t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} −\mathrm{6}{t}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\left({from}\:{here}\:{let}\:{t}\:=\:{t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,{t}_{\mathrm{3}} \:\right) \\ $$$$\:\:{Now} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{3}{y}^{\mathrm{2}} \:=\:\frac{\mathrm{2}}{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{y}} \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{y}}+\frac{\mathrm{2}}{{x}}\:\:=\:\mathrm{2}{q} \\ $$$${since}\:\:{y}={tx} \\ $$$$\:\:\Rightarrow\:\:{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{t}}+\mathrm{1}\right)\:=\:\mathrm{2}{q} \\ $$$$\Rightarrow\:\:\:{x}^{\mathrm{3}} \:=\:\frac{{t}+\mathrm{1}}{{t}^{\mathrm{3}} +{t}} \\ $$$$\Rightarrow\:\:\:\mathrm{2}\boldsymbol{{q}}\:=\:\left(\mathrm{1}+\boldsymbol{{t}}^{\mathrm{2}} \right)\left(\frac{\boldsymbol{{t}}+\mathrm{1}}{\boldsymbol{{t}}^{\mathrm{3}} +\boldsymbol{{t}}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\:\boldsymbol{{q}}=\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }\:\right]^{\mathrm{1}/\mathrm{3}} . \\ $$
Commented by Tinkutara last updated on 06/Oct/18
Isn't there a single answer?
Commented by ajfour last updated on 06/Oct/18
$${something}\:{wrong}\:{with}\:{question}, \\ $$$${it}\:{seems},\:{knowing}\:{it}\:{is}\:{objective} \\ $$$${type}\:{and}\:{roots}\:{are}\:{not}\:{simple}.. \\ $$
Commented by Tinkutara last updated on 06/Oct/18
Yes, thanks Sir.