Value of $$2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1{\text{ is :}}$$

Solution :
We have,
$$\eqalign{ & 2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1 \cr & = 2\left[ {{{\left( {{{\sin }^2}\theta } \right)}^3} + {{\left( {{{\cos }^2}\theta } \right)}^3}} \right] - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1 \cr & = 2\left[ {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\left( {{{\sin }^4}\theta + {{\cos }^4}\theta - {{\sin }^2}\theta \,{{\cos }^2}\theta } \right)} \right] - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right) + 1 \cr & = \left[ {2\,{{\sin }^4}\theta + 2\,{{\cos }^4}\theta - 2\,{{\sin }^2}\theta \,{{\cos }^2}\theta } \right] - 3\,{\sin ^4}\theta - 3\,{\cos ^4}\theta + 1 \cr & = - {\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)^2} + 1 \cr & = - {1^2} + 1 = - 1 + 1 = 0 \cr} $$