Question

(a+b+c)- a3-b3-c3=3(a+b)(b+c)(c+a)

Answer 1

Consider:

(a+b+c)3 - a3-b3-c3 = [(a+b)+c]3 – a3-b3-c3

(a+b)3+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+3ab(a+b)+c3+3c(a+b)(a+b+c)-a3-b3-c3

A3+b3+c3+3ab(a+b)+3c(a+b)(a+b+c)-a3-b3-c3

3ab(a+b)+3c(a+b)(a+b+c) 3(a+b)[ab+c(a+b+c)]

3(a+b)[ab+ac+bc+c2] 3(a+b)[a(b+c)+c(b+c)]

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

Hence proved