Let $${f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)\,$$     where $$x \in R\,\,{\text{and }}k \geqslant 1.$$    Then $${f_4}\left( x \right) - {f_6}\left( x \right)\,$$   equals

Solution :
Let $${f_k}\left( x \right) = \frac{1}{k}\left( {{{\sin }^k}x + {{\cos }^k}x} \right)$$
Consider
$$\eqalign{ & {f_4}\left( x \right) - {f_6}\left( x \right) = \frac{1}{4}\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - \frac{1}{6}\left( {{{\sin }^6}x + {{\cos }^6}x} \right) \cr & = \frac{1}{4}\left[ {1 - 2{{\sin }^2}x{{\cos }^2}x} \right] - \frac{1}{6}\left[ {1 - 3{{\sin }^2}x{{\cos }^2}x} \right] \cr & = \frac{1}{4} - \frac{1}{6} \cr & = \frac{1}{{12}} \cr} $$