if a+b+c=0 then (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b)))=?


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Question Number 8355 by Nayon last updated on 09/Oct/16

$i f a + b + c = 0 t h e n$ $(\frac{a - b}{c} + \frac{b - c}{a} + \frac{c - a}{b}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}) = ?$
Commented by Rasheed Soomro last updated on 11/Oct/16
$=(((a−b)/c))((a/(b−c))+(b/(c−a))+(c/(a−b))) +(((b−c)/a))((a/(b−c))+(b/(c−a))+(c/(a−b))) +(((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b))) ={(((a−b)/c))((a/(b−c))+(b/(c−a)))+1} +{(((b−c)/a))((b/(c−a))+(c/(a−b)))+1} +{(((c−a)/b))((a/(b−c))+(c/(a−b)))+1} =(((a−b)/c))((a/(b−c))+(b/(c−a))) +(((b−c)/a))((b/(c−a))+(c/(a−b))) +(((c−a)/b))((a/(b−c))+(c/(a−b))) +3 =(((a−b)/c))(((ac−a^2 +b^2 −bc)/((b−c)(c−a)))) +(((b−c)/a))(((ab−b^2 +c^2 −ac)/((c−a)(a−b)))) +(((c−a)/b))(((a^2 −ab+bc−c^2 )/((b−c)(a−b)))) +3 =(((a−b)/c))(((−(a−b)(a+b)+c(a−b))/((b−c)(c−a)))) +(((b−c)/a))(((−(b−c)(b+c)+a(b−c))/((c−a)(a−b)))) +(((c−a)/b))(((−(c−a)(c+a)−b(c−a))/((b−c)(a−b)))) +3 =(((a−b)/c))((((a−b)(−a−b+c))/((b−c)(c−a)))) +(((b−c)/a))((((b−c)(−b−c+a))/((c−a)(a−b)))) +(((c−a)/b))((((c−a){−(a+b+c)})/((b−c)(a−b)))) +3 =(((a−b)^2 (−a−b+c))/(c(b−c)(c−a))) +(((b−c)^2 (−b−c+a))/(a(c−a)(a−b))) +(((c−a)^2 {−(0)})/(b(b−c)(a−b))) +3 =(((a−b)^2 (−a−b+c))/(c(b−c)(c−a)))+(((b−c)^2 (a−b−c))/(a(c−a)(a−b)))+3 =((1/(c−a)))((((a−b)^2 (−a−b+c))/(c(b−c)))+(((b−c)^2 (a−b−c))/(a(a−b))))+3 =((1/(c−a)))((((a−b)^2 (−a−b+c)+(b−c)^2 (a−b−c))/(ac(a−b)(b−c))))+3 =((1/(c−a)))((((a−b)^2 (a+b+c−2a−2b)+(b−c)^2 (a+b+c−2b−2c))/(ac(a−b)(b−c))))+3 =((1/(c−a)))((((a−b)^2 (−2a−2b)+(b−c)^2 (−2b−2c))/(ac(a−b)(b−c))))+3 =(((−2)/(c−a)))((((a−b)^2 (a+b)+(b−c)^2 (b+c))/(ac(a−b)(b−c))))+3 Continue to the next comment.$
$= (\frac{a - b}{c}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b})$ $+ (\frac{b - c}{a}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b})$ $+ (\frac{c - a}{b}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b})$ $= {(\frac{a - b}{c}) (\frac{a}{b - c} + \frac{b}{c - a}) + 1}$ $+ {(\frac{b - c}{a}) (\frac{b}{c - a} + \frac{c}{a - b}) + 1}$ $+ {(\frac{c - a}{b}) (\frac{a}{b - c} + \frac{c}{a - b}) + 1}$ $= (\frac{a - b}{c}) (\frac{a}{b - c} + \frac{b}{c - a})$ $+ (\frac{b - c}{a}) (\frac{b}{c - a} + \frac{c}{a - b})$ $+ (\frac{c - a}{b}) (\frac{a}{b - c} + \frac{c}{a - b})$ $+ 3$ $= (\frac{a - b}{c}) (\frac{a c - a^{2} + b^{2} - b c}{(b - c) (c - a)})$ $+ (\frac{b - c}{a}) (\frac{a b - b^{2} + c^{2} - a c}{(c - a) (a - b)})$ $+ (\frac{c - a}{b}) (\frac{a^{2} - a b + b c - c^{2}}{(b - c) (a - b)})$ $+ 3$ $= (\frac{a - b}{c}) (\frac{- (a - b) (a + b) + c (a - b)}{(b - c) (c - a)})$ $+ (\frac{b - c}{a}) (\frac{- (b - c) (b + c) + a (b - c)}{(c - a) (a - b)})$ $+ (\frac{c - a}{b}) (\frac{- (c - a) (c + a) - b (c - a)}{(b - c) (a - b)})$ $+ 3$ $= (\frac{a - b}{c}) (\frac{(a - b) (- a - b + c)}{(b - c) (c - a)})$ $+ (\frac{b - c}{a}) (\frac{(b - c) (- b - c + a)}{(c - a) (a - b)})$ $+ (\frac{c - a}{b}) (\frac{(c - a) {- (a + b + c)}}{(b - c) (a - b)})$ $+ 3$ $= \frac{{(a - b)}^{2} (- a - b + c)}{c (b - c) (c - a)}$ $+ \frac{{(b - c)}^{2} (- b - c + a)}{a (c - a) (a - b)}$ $+ \frac{{(c - a)}^{2} {- (0)}}{b (b - c) (a - b)}$ $+ 3$ $= \frac{{(a - b)}^{2} (- a - b + c)}{c (b - c) (c - a)} + \frac{{(b - c)}^{2} (a - b - c)}{a (c - a) (a - b)} + 3$ $= (\frac{1}{c - a}) (\frac{{(a - b)}^{2} (- a - b + c)}{c (b - c)} + \frac{{(b - c)}^{2} (a - b - c)}{a (a - b)}) + 3$ $= (\frac{1}{c - a}) (\frac{{(a - b)}^{2} (- a - b + c) + {(b - c)}^{2} (a - b - c)}{a c (a - b) (b - c)}) + 3$ $= (\frac{1}{c - a}) (\frac{{(a - b)}^{2} (a + b + c - 2 a - 2 b) + {(b - c)}^{2} (a + b + c - 2 b - 2 c)}{a c (a - b) (b - c)}) + 3$ $= (\frac{1}{c - a}) (\frac{{(a - b)}^{2} (- 2 a - 2 b) + {(b - c)}^{2} (- 2 b - 2 c)}{a c (a - b) (b - c)}) + 3$ $= (\frac{- 2}{c - a}) (\frac{{(a - b)}^{2} (a + b) + {(b - c)}^{2} (b + c)}{a c (a - b) (b - c)}) + 3$ $Continue to the next comment .$
Commented by Rasheed Soomro last updated on 13/Oct/16
$Continue from above a+b+c=0⇒c=−a−b =(((−2)/(c−a)))((((a−b)^2 (a+b)+(b−c)^2 (b+c))/(ac(a−b)(b−c))))+3 =(((−2)/(−a−b−a)))((((a−b)^2 (a+b)+(b−{−a−b})^2 (−a))/(a(−a−b)(a−b)(b−{−a−b}))))+3 =(((−2)/(−2a−b)))((((a−b)^2 (a+b)+(2b+a)^2 (−a))/(a(−a−b)(a−b)(2b+a))))+3 =−((2/(2a+b)))((((a−b)^2 (a+b)−a(2b+a)^2 )/(a(a+b)(a−b)(2b+a))))+3 =−2((((a−b)^2 (a+b)−a(2b+a)^2 )/(a(a+b)(a−b)(2b+a)(2a+b))))+3 =−2((((a^2 −b^2 )(a−b)−a(4b^2 +4ab+a^2 )/(a(a+b)(a−b)(2b+a)(2a+b))))+3 =−2(((−ab^2 −a^2 b+b^3 −4ab^2 −4a^2 b)/(a(a+b)(a−b)(2b+a)(2a+b))))+3 =−2(((−5ab^2 −5a^2 b+b^3 )/(a(a+b)(a−b)(2b+a)(2a+b))))+3$
$Continue from above$ $a + b + c = 0 \Rightarrow c = - a - b$ $= (\frac{- 2}{c - a}) (\frac{{(a - b)}^{2} (a + b) + {(b - c)}^{2} (b + c)}{a c (a - b) (b - c)}) + 3$ $= (\frac{- 2}{- a - b - a}) (\frac{{(a - b)}^{2} (a + b) + {(b - {- a - b})}^{2} (- a)}{a (- a - b) (a - b) (b - {- a - b})}) + 3$ $= (\frac{- 2}{- 2 a - b}) (\frac{{(a - b)}^{2} (a + b) + {(2 b + a)}^{2} (- a)}{a (- a - b) (a - b) (2 b + a)}) + 3$ $= - (\frac{2}{2 a + b}) (\frac{{(a - b)}^{2} (a + b) - a {(2 b + a)}^{2}}{a (a + b) (a - b) (2 b + a)}) + 3$ $= - 2 (\frac{{(a - b)}^{2} (a + b) - a {(2 b + a)}^{2}}{a (a + b) (a - b) (2 b + a) (2 a + b)}) + 3$ $= - 2 (\frac{(a^{2} - b^{2}) (a - b) - a (4 b^{2} + 4 a b + a^{2}}{a (a + b) (a - b) (2 b + a) (2 a + b)}) + 3$ $= - 2 (\frac{- {a b}^{2} - a^{2} b + b^{3} - 4 {a b}^{2} - 4 a^{2} b}{a (a + b) (a - b) (2 b + a) (2 a + b)}) + 3$ $= - 2 (\frac{- 5 {a b}^{2} - 5 a^{2} b + b^{3}}{a (a + b) (a - b) (2 b + a) (2 a + b)}) + 3$
	Answered by Rasheed Soomro last updated on 12/Oct/16

	$a + b + c = 0, (\frac{a - b}{c} + \frac{b - c}{a} + \frac{c - a}{b}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}) = ?$ $- - - - - - - - - - - - - - - - - - - - - - - - - - - -$ $a + b + c = 0 \Rightarrow c = - a - b$ $(\frac{a - b}{c} + \frac{b - c}{a} + \frac{c - a}{b}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b})$ $= (\frac{a - b}{- a - b} + \frac{b - (- a - b)}{a} + \frac{- a - b - a}{b}) (\frac{a}{b - (- a - b)} + \frac{b}{- a - b - a} + \frac{- a - b}{a - b})$ $= (- \frac{a - b}{a + b} + \frac{a + 2 b}{a} - \frac{2 a + b}{b}) (\frac{a}{a + 2 b} - \frac{b}{2 a + b} - \frac{a + b}{a - b})$ $= (- \frac{a - b}{a + b}) (\frac{a}{a + 2 b} - \frac{b}{2 a + b} - \frac{a + b}{a - b})$ $+ (\frac{a + 2 b}{a}) (\frac{a}{a + 2 b} - \frac{b}{2 a + b} - \frac{a + b}{a - b})$ $+ (- \frac{2 a + b}{b}) (\frac{a}{a + 2 b} - \frac{b}{2 a + b} - \frac{a + b}{a - b})$ $= (- \frac{a (a - b)}{(a + b) (a + 2 b)} + \frac{b (a - b)}{(a + b) (2 a + b)} + 1)$ $+ (1 - \frac{b (a + 2 b)}{a (2 a + b)} - \frac{(a + b) (a + 2 b)}{a (a - b)})$ $+ (- \frac{a (2 a + b)}{b (a + 2 b} + 1 + \frac{(a + b) (2 a + b)}{b (a - b)})$ $= - \frac{a (a - b)}{(a + b) (a + 2 b)} + \frac{b (a - b)}{(a + b) (2 a + b)}$ $- \frac{b (a + 2 b)}{a (2 a + b)} - \frac{(a + b) (a + 2 b)}{a (a - b)}$ $- \frac{a (2 a + b)}{b (a + 2 b} + \frac{(a + b) (2 a + b)}{b (a - b)}$ $+ 3$ $=$ $C ontinue$
	Answered by Rasheed Soomro last updated on 11/Oct/16

	$a + b + c = 0, (\frac{a - b}{c} + \frac{b - c}{a} + \frac{c - a}{b}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b}) = ?$ $(\frac{a - b}{c} + \frac{b - c}{a} + \frac{c - a}{b}) (\frac{a}{b - c} + \frac{b}{c - a} + \frac{c}{a - b})$ $= (\frac{a + b + c - 2 b - c}{c} + \frac{a + b + c - a - 2 c}{a} + \frac{a + b + c - 2 a - b}{b})$ $\times (\frac{a}{a + b + c - a - 2 c} + \frac{b}{a + b + c - 2 a - b} + \frac{c}{a + b + c - 2 b - c})$ $= (\frac{- 2 b - c}{c} + \frac{- a - 2 c}{a} + \frac{- 2 a - b}{b})$ $\times (\frac{a}{- a - 2 c} + \frac{b}{- 2 a - b} + \frac{c}{- 2 b - c})$ $a + b + c = 0 \Rightarrow c = - a - b$ $= (\frac{- 2 b - (- a - b)}{- a - b} + \frac{- a - 2 (- a - b)}{a} + \frac{- 2 a - b}{b})$ $\times (\frac{a}{- a - 2 (- a - b)} + \frac{b}{- 2 a - b} + \frac{- a - b}{- 2 b - (- a - b})$ $= (\frac{- 2 b + a + b)}{- a - b} + \frac{- a + 2 a + 2 b)}{a} + \frac{- 2 a - b}{b})$ $\times (\frac{a}{- a + 2 a + 2 b)} + \frac{b}{- 2 a - b} + \frac{- a - b}{- 2 b + a + b})$ $= (\frac{a - b}{- a - b} + 1 + \frac{a + 2 b}{a} + 1 + \frac{- 2 a - b}{b} + 1 - 3)$ $\times (\frac{a}{a + 2 b} + 1 + \frac{b}{- 2 a - b} + 1 + \frac{- a - b}{a - b} + 1 - 3)$ $= (\frac{a - b}{- a - b} + 1 + \frac{a + 2 b}{a} + 1 + \frac{- 2 a - b}{b} + 1 - 3)$ $\times (\frac{a}{a + 2 b} + 1 + \frac{b}{- 2 a - b} + 1 + \frac{- a - b}{a - b} + 1 - 3)$ $= (\frac{- 2 b}{- a - b} + \frac{2 a + 2 b}{a} + \frac{- 2 a}{b} - 3)$ $\times (\frac{2 a + 2 b}{a + 2 b} + \frac{- 2 a}{- 2 a - b} + \frac{- 2 b}{a - b} - 3)$ $Continue$
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