$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$ is equal to
(a) 0
(b) 1
(c) −1
(d) None of these
The given expression is $2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$.
Simplifying the given expression, we have
$2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)-3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)$
$=2 \sin ^{6} \theta+2 \cos ^{6} \theta-3 \sin ^{4} \theta-3 \cos ^{4} \theta$
$=\left(2 \sin ^{6} \theta-3 \sin ^{4} \theta\right)+\left(2 \cos ^{6} \theta-3 \cos ^{4} \theta\right)$
$=\sin ^{4} \theta\left(2 \sin ^{2} \theta-3\right)+\cos ^{4} \theta\left(2 \cos ^{2} \theta-3\right)$
$=\sin ^{4} \theta\left\{2\left(1-\cos ^{2} \theta\right)-3\right\}+\cos ^{4} \theta\left\{2\left(1-\sin ^{2} \theta\right)-3\right\}$
$=\sin ^{4} \theta\left(2-2 \cos ^{2} \theta-3\right)+\cos ^{4} \theta\left(2-2 \sin ^{2} \theta-3\right)$
$=\sin ^{4} \theta\left(-1-2 \cos ^{2} \theta\right)+\cos ^{4} \theta\left(-1-2 \sin ^{2} \theta\right)$
$=-\sin ^{4} \theta-2 \sin ^{4} \theta \cos ^{2} \theta-\cos ^{4} \theta-2 \cos ^{4} \theta \sin ^{2} \theta$
$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{4} \theta \sin ^{2} \theta-2 \sin ^{4} \theta \cos ^{2} \theta$
$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{2} \theta \sin ^{2} \theta\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$
$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{2} \theta \sin ^{2} \theta(1)$
$=-\sin ^{4} \theta-\cos ^{4} \theta-2 \cos ^{2} \theta \sin ^{2} \theta$
$=-\left(\sin ^{4} \theta+\cos ^{4} \theta+2 \cos ^{2} \theta \sin ^{2} \theta\right)$
$=-\left\{\left(\sin ^{2} \theta\right)^{2}+\left(\cos ^{2} \theta\right)^{2}+2 \sin ^{2} \theta \cos ^{2} \theta\right\}$
$=-\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}$
$=-(1)^{2}$
$=-1$
Therefore, the correct option is (c).