Question

prove that (a-b)^2+(b-c)^2+(c-a)^2=2(a-b)(a-c)+2(b-c)(b-a)+2(c-a)(c-b)

Answer 1

given,

RHS=2(a-b)(a-c)+2(b-c)(b-a)+2(c-a)(c-b)

=2(a²-ac-ab+bc)+2(b²-ab-bc+ac)+2(c²-bc-ac+ab)

=2a²-2ac-2ab+2bc+2b²-2ab-2bc+2ac+2c²-2bc-2ac+2ab

=2a²+2b²+2c²-2ab-2ab+2ab+2bc-2bc-2bc
-2ac+2ac-2ac

=2a²+2b²+2c²-2ab-2bc-2ac

=a²+a²+b²+b²+c²+c²-2ab-2bc-2ac

=a²+b²-2ab+b²+c²-2bc+c²+a²-2ac

=(a-b)²+(b-c)²+(c-a)²

=LHS

LHS=RHS

hence proved

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