Question

prove that:(a+b)3+(b+c)3+(c+a)3-3(a+b) (b+c) (c+a)=2(a3+b3+c3-3abc)​

Answer 1

Step-by-step explanation:

LHS;

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

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

=1 (cancellation)

RHS:

2(a3+b3+c3-3abc)

=a6+b6+c6-6abc

=1 (by cancellation)

;LHS=RHS