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Question Number 44876 by Tinkutara last updated on 05/Oct/18

	Answered by ajfour last updated on 06/Oct/18
	$let z_1 = x , z_2 = y (just calling) { (( x(x^2 −3y^2 )=2)),(( y(3x^2 −y^2 )=1)) :} Then x^3 −3xy^2 = 6x^2 y−2y^3 let y=tx 1−3t^2 = 6t−2t^3 or 2t^3 −3t^2 −6t+1 = 0 (from here let t = t_1 , t_2 ,t_3 ) Now x^2 −3y^2 = (2/x) 3x^2 −y^2 = (1/y) ⇒ 2x^2 +2y^2 = (1/y)+(2/x) = 2q since y=tx ⇒ x^2 (1+t^2 )=(1/x)((1/t)+1) = 2q ⇒ x^3 = ((t+1)/(t^3 +t)) ⇒ 2q = (1+t^2 )(((t+1)/(t^3 +t)))^(2/3) q=(x^2 +y^2 )= (1/2)[ (((1+t^2 )(1+t)^2 )/t^2 ) ]^(1/3) .$
	$l e t z_{1} = x, z_{2} = y (j u s t c a l l i n g)$ ${\begin{cases} x (x^{2} - 3 y^{2}) = 2 \\ y (3 x^{2} - y^{2}) = 1 \end{cases}$ $T h e n x^{3} - 3 {x y}^{2} = 6 x^{2} y - 2 y^{3}$ $l e t y = t x$ $1 - 3 t^{2} = 6 t - 2 t^{3}$ $o r 2 t^{3} - 3 t^{2} - 6 t + 1 = 0$ $(f r o m h e r e l e t t = t_{1}, t_{2}, t_{3})$ $N o w$ $x^{2} - 3 y^{2} = \frac{2}{x}$ $3 x^{2} - y^{2} = \frac{1}{y}$ $\Rightarrow 2 x^{2} + 2 y^{2} = \frac{1}{y} + \frac{2}{x} = 2 q$ $s i n c e y = t x$ $\Rightarrow x^{2} (1 + t^{2}) = \frac{1}{x} (\frac{1}{t} + 1) = 2 q$ $\Rightarrow x^{3} = \frac{t + 1}{t^{3} + t}$ $\Rightarrow 2 q = (1 + t^{2}) {(\frac{t + 1}{t^{3} + t})}^{2 / 3}$ $q = (x^{2} + y^{2}) = \frac{1}{2} {[\frac{(1 + t^{2}) {(1 + t)}^{2}}{t^{2}}]}^{1 / 3} .$
	Commented by Tinkutara last updated on 06/Oct/18
	Isn't there a single answer?
	Commented by ajfour last updated on 06/Oct/18

	$s o m e t h i n g w r o n g w i t h q u e s t i o n,$ $i t s e e m s, k n o w i n g i t i s o b j e c t i v e$ $t y p e a n d r o o t s a r e n o t s i m p l e . .$
	Commented by Tinkutara last updated on 06/Oct/18
	Yes, thanks Sir.
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