Three real numbers a,b,c satisfying ab+c=10,bc+a=11,ca+b=14. Find (a−b)(b−c)(c−a)(a−1)(b−1)(c−1)


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Question Number 110596 by Aina Samuel Temidayo last updated on 29/Aug/20

$Three real numbers a, b, c satisfying$ $ab + c = 10, bc + a = 11, ca + b = 14 . Find$ $(a - b) (b - c) (c - a) (a - 1) (b - 1) (c - 1)$
Commented by Her_Majesty last updated on 29/Aug/20

$o n e s o l u t i o n i s$ $a = 3 b = 2 c = 4 \Rightarrow a n s w e r i s - 12$
Commented by Aina Samuel Temidayo last updated on 29/Aug/20

$Thanks but I {don}^{'} t only need answers .$ $I also need well detailed solutions .$
	Answered by Her_Majesty last updated on 30/Aug/20

	$a b + c = 10$ $b c + a = 11$ $c a + b = 14$ $=====$ $c = 10 - a b = \frac{11 - a}{b} = \frac{14 - b}{a}$ $10 - a b = \frac{14 - b}{a} \Rightarrow b = \frac{2 (5 a - 7)}{a^{2} - 1}$ $c = \frac{2 (7 a - 5)}{a^{2} - 1} = - \frac{(a - 11) (a^{2} - 1)}{2 (5 a - 7)}$ $\frac{2 (7 a - 5)}{a^{2} - 1} = - \frac{(a - 11) (a^{2} - 1)}{2 (5 a - 7)}$ $a^{5} - 11 a^{4} - 2 a^{3} + 162 a^{2} - 295 a + 129 = 0$ $t r y i n g f a c t o r s o f 129 w h i c h a r e \pm 1, \pm 3, \pm 43, \pm 129$ $\Rightarrow a_{1} = 3$ $\Rightarrow b_{1} = 2, c_{1} = 4$ $n o w w e m u s t s o l v e t h i s :$ $a^{4} - 8 a^{3} - 26 a^{2} + 84 a - 43 = 0$ $s a d l y {t h e r e}^{'} s n o o t h e r e x a c t s o l u t i o n$ $a_{2} \approx - 4.07488$ $a_{3} \approx . 685287$ $a_{4} \approx 1.56780$ $a_{5} \approx 9.82180$ $a n y w a y a l l 5 t r i p l e t s (a / b / c) g i v e - 12$
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