Question

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

Answer 1

Answer:

a3 + b3 + c3 - 3abc = (a + b + c )(a2 + b2 + c2 -ab - ac -bc) Now it is given that : a3 + b3 + c3 = 3abc. So, a3 + b3 + c3 - 3abc = 0.

Answer 2

Answer:

a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3 = [a(b-c)]^3 + [b(c-a)]^3 + [c(a-b)]^3

We know, a^3 + b^3 + c^3 =3abc if a+b+c =0

Here, a =[a(b-c)] , b = [b(c-a)] & c =[c(a-b)]

⇒a+b+c =  [a(b-c)] + [b(c-a)] + [c(a-b)]

             = ab - ac + bc - ab + ac - bc = 0

Therefore, a^3 + b^3 + c^3 = 3abc

⇒[a(b-c)]^3 + [b(c-a)]^3 + [c(a-b)]^3 = 3 [a(b-c)] [b(c-a)] [c(a-b)]

                                                     = 3abc* (b-c)* (c-a)* (a-b)