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Question Number 181138 by Agnibhoo98 last updated on 22/Nov/22

If a + b + c = 0 then, (1/(2a^2 + bc)) + (1/(2b^2 + ca)) + (1/(2c^2 + ab)) = ?

Ifa+b+c=0then,12a2+bc+12b2+ca+12c2+ab=?

Answered by som(math1967) last updated on 22/Nov/22

a+b+c=0 ⇒a^2 =−ab−ca b^2 =−ab−bc c^2 =−ca−bc 2a^2 +bc=a^2 −ab−ca+bc=(a−b)(a−c) 2b^2 +ca=(b−c)(b−a) 2c^2 +ab=(c−a)(c−b) (1/(2a^2 + bc)) + (1/(2b^2 + ca)) + (1/(2c^2 + ab)) =(1/((a−b)(a−c)))+(1/((b−c)(b−a)))+(1/((c−a)(c−b))) =−(1/((a−b)(c−a)))−(1/((b−c)(a−b)))−(1/((c−a)(b−c))) =−[((b−c+c−a+a−b)/((a−b)(b−c)(c−a)))]=0

a+b+c=0⇒a2=−ab−cab2=−ab−bcc2=−ca−bc2a2+bc=a2−ab−ca+bc=(a−b)(a−c)2b2+ca=(b−c)(b−a)2c2+ab=(c−a)(c−b)12a2+bc+12b2+ca+12c2+ab=1(a−b)(a−c)+1(b−c)(b−a)+1(c−a)(c−b)=−1(a−b)(c−a)−1(b−c)(a−b)−1(c−a)(b−c)=−[b−c+c−a+a−b(a−b)(b−c)(c−a)]=0

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