Factorise each of the following:
We know that(x + y)3 = x3 + y3 + 3xy(x + y)| Using Identity VI⇒ x3 + y3 = (x + y)3 - 3xy(x + y)⇒ x3 + y3 = (x + y){(x + y)2 - 3xy)⇒ x3 + y3 = (x + y)(x2 + 2xy + y2 - 3xy)| Using Identity I⇒ x3 + y3 = (x + y)(x2 - xy + y2)