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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(a−b)(a−c)+b2(b−c)(b−a)+c2(c−a)(c−b)=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(a−b)(a−c)+b2(b−c)(b−a)+c2(c−a)(c−b)=a+b+cLHS−a2(b−c)−b2(c−a)−c2(a−b)(a−b)(b−c)(c−a)=−a2b−ab2+b2c−a2c−bc2+ac2(a−b)(b−c)(c−a)=−ab(a−b)+c(b2−a2)−c2(b−a)(a−b)(b−c)(c−a)=−ab(a−b)−c(a2−b2)+c2(a−b)(a−b)(b−c)(c−a)=−(a−b){ab−c(a+b)+c2}(a−b)(b−c)(c−a)=−(a−b){ab−ac−bc+c2}(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

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