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Question Number 30456 by daffa123 last updated on 22/Feb/18
proof that  (a^2 /((a−b)(a−c)))+(b^2 /((b−c)(b−a)))+(c^2 /((c−a)(c−b)))= a+b+c
proofthata2(ab)(ac)+b2(bc)(ba)+c2(ca)(cb)=a+b+c
Answered by Rasheed.Sindhi last updated on 22/Feb/18
(a^2 /((a−b)(a−c)))+(b^2 /((b−c)(b−a)))+(c^2 /((c−a)(c−b)))= a+b+c  LHS  ((−a^2 (b−c)−b^2 (c−a)−c^2 (a−b))/((a−b)(b−c)(c−a)))  =−((a^2 b−ab^2 +b^2 c−a^2 c−bc^2 +ac^2 )/((a−b)(b−c)(c−a)))  =−((ab(a−b)+c(b^2 −a^2 )−c^2 (b−a))/((a−b)(b−c)(c−a)))  =−((ab(a−b)−c(a^2 −b^2 )+c^2 (a−b))/((a−b)(b−c)(c−a)))  =−(((a−b){ab−c(a+b)+c^2 })/((a−b)(b−c)(c−a)))  =−(((a−b){ab−ac−bc+c^2 })/((a−b)(b−c)(c−a)))  =−((a(b−c)−c(b−c))/((b−c)(c−a)))  =−(((b−c)(a−c))/((b−c)(c−a)))  =(((b−c)(c−a))/((b−c)(c−a)))  =1≠a+b+c
a2(ab)(ac)+b2(bc)(ba)+c2(ca)(cb)=a+b+cLHSa2(bc)b2(ca)c2(ab)(ab)(bc)(ca)=a2bab2+b2ca2cbc2+ac2(ab)(bc)(ca)=ab(ab)+c(b2a2)c2(ba)(ab)(bc)(ca)=ab(ab)c(a2b2)+c2(ab)(ab)(bc)(ca)=(ab){abc(a+b)+c2}(ab)(bc)(ca)=(ab){abacbc+c2}(ab)(bc)(ca)=a(bc)c(bc)(bc)(ca)=(bc)(ac)(bc)(ca)=(bc)(ca)(bc)(ca)=1a+b+c

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