Question Number 11476 by tawa last updated on 26/Mar/17

Answered by sandy_suhendra last updated on 27/Mar/17

(1/((a−b)(b−c))) + (1/((b−c)(c−a))) + (1/((c−a)(a−b))) =((c−a+a−b+b−c)/((a−b)(b−c)(c−a))) = 0 (1/((a−b)^2 )) + (1/((b−c)^2 )) + (1/((c−a)^2 )) =[(1/((a−b))) + (1/((b−c))) + (1/((c−a)))]^2 − 2[(1/((a−b)(b−c)))+(1/((b−c)(c−a)))+(1/((c−a)(a−b)))] =[(1/((a−b)))+(1/((b−c)))+(1/((c−a)))]^2 − 2.0 =[(1/((a−b)))+(1/((b−c)))+(1/((c−a)))]^2

Commented bytawa last updated on 27/Mar/17

God bless you sir.