(b) $3\left\{ {{{\sin }^4}\left( {\frac{{3\pi }}{2} – \alpha } \right) + {{\sin }^4}(3\pi + \alpha )} \right\}$
$ – 2\left\{ {{{\sin }^6}\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi – \alpha )} \right\}$
$ = 3\left\{ {\,{{( – \cos \alpha )}^4} + {{( – \sin \alpha )}^4}} \right\} – 2\,\left\{ {{{\cos }^6}\alpha + {{\sin }^6}\alpha } \right\}$
$ = 3\left\{ {{{\cos }^2}\alpha + {{\sin }^2}\alpha {)^2} – 2{{\sin }^2}\alpha {{\cos }^2}\alpha } \right\}$
$ – 2\left\{ {{{({{\cos }^2}\alpha + {{\sin }^2}\alpha )}^3} – 3{{\cos }^2}\alpha {{\sin }^2}\alpha ({{\cos }^2}\alpha + {{\sin }^2}\alpha )} \right\}$
$ = 3 – 6{\sin ^2}\alpha {\cos ^2}\alpha – 2 + 6{\sin ^2}\alpha {\cos ^2}\alpha = 3 – 2 = 1$
Trick : Put $\alpha = 0,\frac{\pi }{2},$ the value of expression remains $1$ i.e., it is independent of $\alpha $.