The value of the following is : $3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+$ $2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)+12 \sin ^{2} \theta \cos ^{2} \theta$

Expression, $=3\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+2\left(\sin ^{6} \theta+\cos ^{6} \theta\right)+12 \sin ^{2}$ $\theta \cdot \cos ^{2} \theta$ $=3\left\{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cdot \cos ^{2} \theta\right\}$ $+2\left\{\left(\sin ^{2} \theta+\cos ^{2} \theta\right) 3-3 \sin ^{2} \theta \cdot \cos ^{2} \theta\left(\sin ^{2} \theta+\right.\right.$ $\left.\cos ^{2} \theta\right)+12 \sin ^{2} \theta \cdot \cos ^{2} \theta$ $\left[\because a^{2}+b^{2}=(a+b)^{2}-2 a b ; a^{3}+b^{3}=(a+b)^{3}-\right.$ $3 a b(a+b)]$ $=3\left(1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta\right)+2\left(1-3 \sin ^{2} \theta \cdot \cos ^{2} \theta\right)$ $+12 \sin ^{2} \theta \cdot \cos ^{2} \theta=3-6 \sin ^{2} \theta \cdot \cos ^{2} \theta+2-6$ $\sin ^{2} \theta \cos ^{2} \theta+12 \sin ^{2} \theta \cdot \cos ^{2} \theta=5$

Alternative:

Let $\theta=0^{\circ}$ $\Rightarrow 3\left(\sin ^{4} 0^{\circ}+\cos ^{4} 0^{\circ}\right)+2\left(\sin ^{6} 0^{\circ}+\cos ^{6} 0^{\circ}\right)+12$ $\sin ^{2} 0^{\circ} \cdot \cos ^{2} 0^{\circ}=3(1+0)+2(1+0)+0=5$