Question

Prove that (a+b+c) 3 -a 3 -b 3 -c 3 = 3(a+b) (b+c) (c+a)

Answer 1

(a+b+c)^3-a^3-b^3-C^3= a^3+b^3+c^3+3(a+b)(b+c)(a+c)-a^3-b^3-c^3
= 3(a+b)(b+c)(c+a) proved

Answer 2

(a+b+c)^3 = {a+(b+c)}^3 = a^3 +(b+c)^3 + 3a(b+c){a+(b+c)} 
= a^3 +(b+c)^3 +3(ab+ac)(a+b+c)  [ using (a + b)^3 = a^3 +b^3+3ab(a+b) ]
= a^3 +[b^3+c^3+3bc(b+c)]+3(ab+ac)(a+b+c) 
=(a^3+b^3+c^3) + 3bc(b+c) +3(ab+ac)(a+b+c) 
i.e. (a+b+c)^3 -(a^3+b^3+c^3)= 3bc(b+c)+3(ab+ac)(a+b+c) 
                                      = 3b^2c+3bc^2 +3(a^2b+ab^2+abc+a^2c+abc+ac^2) 
                                       =3[b^2c+bc^2+a^2b+ab^2+a^2c+ac^2+abc+abc ]
                                      =3[(abc+ac^2+b^2c+bc^2) + (a^2b+a^2c+ab^2+abc)] 
                                       =3[c(ab+ac+b^2+bc)+ a(ab+ac+b^2+bc)] 
                                       = 3[(ab+ac+b^2+bc)(c+a)] 
                                       =3[{a(b+c)+b(b+c)}(c+a)] 
                                       =3[{(a+b)(b+c)}(c+a)] 
                                       =3(a+b)(b+c)(c+a) 
                    Hence LHS = RHS proved