Biểu thức $f\left( x \right) = 3\left( {{{\sin }^4}x + {{\cos }^4}x} \right) - 2\left( {{{\sin }^6}x + {{\cos }^6}x} \right)$ có giá trị bằng:

${\sin ^4}x + {\cos ^4}x ={\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x\\= 1 - 2{\sin ^2}x{\cos ^2}x$

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

$f\left( x \right) = 3\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right) - 2\left( {1 - 3{{\sin }^2}x{{\cos }^2}x} \right) = 1$