The value of $6\left({\sin }^6\mathrm{\theta }+{\cos }^6\mathrm{\theta }\right)-9\left({\sin }^4\mathrm{\theta }+{\cos }^4\mathrm{\theta }\right)+4$is

Solution:

$\begin{array}{l}6\left({\sin }^6\mathrm{\theta }+{\cos }^6\mathrm{\theta }\right)-9\left({\sin }^4\mathrm{\theta }+{\cos }^4\mathrm{\theta }\right)+4\\ =6\left[{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)}^3-3{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)\right]-9\left[{\left({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)}^2-2{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right]+4\\ \begin{array}{l}=6\left[1-3{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right]-9\left(1-2{\sin }^2{\mathrm{\theta cos}}^2\mathrm{\theta }\right)+4\\ =6-9+4=1\end{array}\end{array}$