let a,b,c,x,y and z be complex number such that a=((b+c)/(x−2)) ,b=((c+a)/(y−2)) c=((a+b)/(z−2)). xy +yz +zx=1000 and x+y+z=2016 find the value of xyz.


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Question Number 25066 by Mr easy last updated on 02/Dec/17

$l e t a, b, c, x, y a n d z b e c o m p l e x n u m b e r$ $s u c h t h a t a = \frac{b + c}{x - 2}, b = \frac{c + a}{y - 2} c = \frac{a + b}{z - 2} .$ $x y + y z + z x = 1000 a n d x + y + z = 2016$ $f i n d t h e v a l u e o f x y z .$
	Answered by ajfour last updated on 03/Dec/17

	$Π (x - 2) = x y z - 2 Σ x y + 4 Σ x - 8$ $\frac{(a + b) (b + c) (c + a)}{a b c} = x y z - 2000$ $+ 8064 - 8$ $\Rightarrow x y z = \frac{(a + b) (b + c) (c + a)}{a b c} - 6056$ $F u r t h e r,$ $x + y + z = 8 + Σ (\frac{b + c}{a}) = 2016 . . . (i)$ $Σ (x - 2) (y - 2) = Σ (\frac{b + c}{a}) (\frac{c + a}{b})$ $Σ x y - 2 Σ (x + y) + 12 = Σ (\frac{b + c}{a}) (\frac{c + a}{b})$ $1000 - 4 \times 2016 + 12 = Σ (\frac{b + c}{a}) (\frac{c + a}{b})$ $\Rightarrow Σ (\frac{b + c}{a}) (\frac{c + a}{b}) = 7056$ $s o \frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c} a r e r o o t s o f$ $c u b i c e q u a t i o n :$ $q^{3} - 2008 q^{2} + 7056 q - (x y z + 6056) = 0$ $. . . . m a y c o n t i n u e . .$
	Commented by jota+ last updated on 03/Dec/17

	$T h e e n d i s$ $x y z = \frac{(a + b) (b + c) (c + a)}{a b c} - 6056 .$
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