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Question Number 50547 by Pk1167156@gmail.com last updated on 17/Dec/18

	Answered by behi83417@gmail.com last updated on 17/Dec/18
	$35+2(xy+yz+zx)=1 ⇒xy+yz+zx=−17⇒xy=−17−z(y+x) (x+y)(x^2 +y^2 −xy)+z^3 =97 (1−z)(35−z^2 −(−17−z(1−z))=97 (1−z)(52+z−2z^2 )=97 52+z−2z^2 −52z−z^2 +2z^3 =97 ⇒2z^3 −3z^2 −51z−45=0 z=−3.72,−0.97,+6.2 ⇒1) { ((z=−3.72⇒x+y=4.72)),((xy=−17−(−3.72)(4.72)=0.56)) :} t^2 −4.72t+0.56=0⇒t=((4.72±4.48)/2) ⇒(x,y,z)=(4.6,0.12,−3.72),(0.12,4.6,−3.72) 2) { ((z=−.97⇒x+y=1.97)),((xy=−17−(−.97)(1.97)=−15.09)) :} t^2 −1.97t−15.09=0⇒t=((1.97±8.02)/2) ⇒(x,y,z)#(5,−3,−1),(−3,5,−1) 3) { ((z=6.2⇒x+y=−5.2)),((xy=−17−(6.2)(−5.2)=15.24)) :} t^2 +5.2t+15.24=0⇒t=((−5.2±5.82i)/2) ⇒(x,y,z)=(−2.6+2.91i,−2.6−2.91i,6.2), ,(−2.6−2.91i,−2.6+2.91i,6.2) .■$
	$35 + 2 (x y + y z + z x) = 1$ $\Rightarrow x y + y z + z x = - 17 \Rightarrow x y = - 17 - z (y + x)$ $(x + y) (x^{2} + y^{2} - x y) + z^{3} = 97$ $(1 - z) (35 - z^{2} - (- 17 - z (1 - z)) = 97$ $(1 - z) (52 + z - 2 z^{2}) = 97$ $52 + z - 2 z^{2} - 52 z - z^{2} + 2 z^{3} = 97$ $\Rightarrow 2 z^{3} - 3 z^{2} - 51 z - 45 = 0$ $z = - 3.72, - 0.97, + 6.2$ $\Rightarrow 1) {\begin{cases} z = - 3.72 \Rightarrow x + y = 4.72 \\ x y = - 17 - (- 3.72) (4.72) = 0.56 \end{cases}$ $t^{2} - 4.72 t + 0.56 = 0 \Rightarrow t = \frac{4.72 \pm 4.48}{2}$ $\Rightarrow (x, y, z) = (4.6, 0.12, - 3.72), (0.12, 4.6, - 3.72)$ $2) {\begin{cases} z = - . 97 \Rightarrow x + y = 1.97 \\ x y = - 17 - (- . 97) (1.97) = - 15.09 \end{cases}$ $t^{2} - 1.97 t - 15.09 = 0 \Rightarrow t = \frac{1.97 \pm 8.02}{2}$ $You can't use 'macro parameter character #' in math mode$ $3) {\begin{cases} z = 6.2 \Rightarrow x + y = - 5.2 \\ x y = - 17 - (6.2) (- 5.2) = 15.24 \end{cases}$ $t^{2} + 5.2 t + 15.24 = 0 \Rightarrow t = \frac{- 5.2 \pm 5.82 i}{2}$ $\Rightarrow (x, y, z) = (- 2.6 + 2.91 i, - 2.6 - 2.91 i, 6.2),$ $, (- 2.6 - 2.91 i, - 2.6 + 2.91 i, 6.2) . ◼$
	Commented by Pk1167156@gmail.com last updated on 18/Dec/18
	thank you!
	Answered by mr W last updated on 17/Dec/18

	$\Rightarrow x + y + z = 1$ ${(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x)$ $1 = 35 + 2 (x y + y z + z x)$ $\Rightarrow x y + y z + z x = - 17$ ${(x + y + z)}^{3} = x^{3} + y^{3} + z^{3} + 3 (x y + y z + z x) (x + y + z) - 3 x y z$ $1 = 97 + 3 (- 17) (1) - 3 x y z$ $\Rightarrow x y z = 15$ $x, y, z a r e r o o t s o f e q n .$ $p^{3} - p^{2} - 17 p - 15 = 0$ $(p - 5) (p + 1) (p + 3) = 0$ $p = 5, - 1, - 3$ $\Rightarrow x, y, z \in (5, - 1, - 3)$ $====$ $g e n e r a l l y$ $x + y + z = a$ $x^{2} + y^{2} + z^{2} = b$ $x^{3} + y^{3} + z^{3} = c$ $\Rightarrow x y + y z + z x = \frac{a^{2} - b}{2}$ $\Rightarrow x y z = \frac{a^{3} - 3 a b + 2 c}{6}$ $x, y, z a r e r o o t s o f e q n .$ $p^{3} - {a p}^{2} + \frac{a^{2} - b}{2} p - \frac{a^{3} - 3 a b + 2 c}{6} = 0$
	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

	$e x c e l l e n t . . . .$
	Commented by afachri last updated on 17/Dec/18

	$Amazing . Inpired thinker you are, Sir .$
	Commented by OTCHRRE ABDULLAI last updated on 17/Dec/18

	$M y g r e a t m a n t h a t$
	Commented by Necxx last updated on 17/Dec/18

	$i p o s t e d t h i s s a m e q u e s t i o n 2 y e a r s$ $a n d m r W s o l v e d i t . T o m y g r e a t e s t$ $s u r p r i s e t h e s t y l e h e u s e d i s$ $t o t a l l y d i f f e r e n t f r o m n o w .$ $M r W i s i n d e e d a d e e p t h i n k e r .$
	Commented by Pk1167156@gmail.com last updated on 18/Dec/18
	great sir!
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