Question

prove that
(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)

Answer 1

Step-by-step explanation:

LHS =(a-b)3+(b-c)3+(c-a)3

=3a-3b+3b-3c+3c-3a

=0

RHS =3 (a-b)(b-c)(c-a)

=3 ((ab-ac-b^2+bc)(c-a))

=3 (abc-a^2b-ac^2+a^2c-cb^2+ab^2+bc^2-abc)

=3 (0)

=0

Hence, LHS=RHS

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