solve for x,y,z∈R. x^2 +yz=1 y^2 +xz=2 z^2 +xy=3


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Question Number 49188 by behi83417@gmail.com last updated on 04/Dec/18

$s o l v e f o r x, y, z \in R .$ $x^{2} + yz = 1$ $y^{2} + xz = 2$ $z^{2} + xy = 3$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

$e x c e l l e n t p r o b l e m . . . s t i l l t o f i n d t h e g o a l o f$ $x y z . . .$
	Answered by ajfour last updated on 04/Dec/18

	$l e t y = p x, z = q x$ $\Rightarrow x^{2} (1 + p q) = 1$ $x^{2} (p^{2} + q) = 2$ $x^{2} (q^{2} + p) = 3$ $i f p q \neq - 1 t h e n$ $p^{2} + q = 2 + 2 p q$ $q^{2} + p = 3 + 3 p q$ $\Rightarrow q = \frac{p^{2} - 2}{2 p - 1}$ $\Rightarrow {(\frac{p^{2} - 2}{2 p - 1})}^{2} + p = 3 + 3 p (\frac{p^{2} - 2}{2 p - 1})$ $i f p \neq \frac{1}{2}, t h e n$ ${(p^{2} - 2)}^{2} + (p - 3) {(2 p - 1)}^{2} = 3 p (2 p - 1) (p^{2} - 2)$ $\Rightarrow$ $p^{4} - 4 p^{2} + 4 + 4 p^{3} - 16 p^{2} + 13 p - 3$ $- 6 p^{4} + 3 p^{3} + 12 p^{2} - 6 p = 0$ $\Rightarrow 5 p^{4} - 7 p^{3} + 8 p^{2} - 7 p - 1 = 0$ $. . . . .$
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