Answer
$${\sin ^3}\theta + {\cos ^3}\theta = \left( {\cos \theta + \sin \theta } \right)\left( {1 - \cos \theta \sin \theta } \right)$$
Work Step by Step
$$\eqalign{
& {\sin ^3}\theta + {\cos ^3}\theta = \left( {\cos \theta + \sin \theta } \right)\left( {1 - \cos \theta \sin \theta } \right) \cr
& {\text{We transform the more complicated left side to match the right side}}. \cr
& {\text{Recall that }}{a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right),{\text{ then}} \cr
& {\sin ^3}\theta + {\cos ^3}\theta = \left( {\sin \theta + \cos \theta } \right)\left( {{{\sin }^2}\theta - \sin \theta \cos \theta + {{\cos }^2}\theta } \right) \cr
& {\sin ^3}\theta + {\cos ^3}\theta = \left( {\cos \theta + \sin \theta } \right)\left( {{{\cos }^2}\theta + {{\sin }^2}\theta - \sin \theta \cos \theta } \right) \cr
& {\sin ^3}\theta + {\cos ^3}\theta = \left( {\cos \theta + \sin \theta } \right)\left( {1 - \sin \theta \cos \theta } \right) \cr
& {\text{Thus have verified that the given equation is an identity}} \cr} $$
