Given: $10 \sin ^4 \alpha+15 \cos ^4 \alpha=6$
$\begin{aligned} & \Rightarrow 10\left(\sin ^4 \alpha+\cos ^4 \alpha\right)+5 \cos ^4 \alpha=6 \\ & \Rightarrow 10\left(\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2-2 \sin ^2 \alpha \cdot \cos ^2 \alpha\right)=6-5 \cos ^4 \alpha \\ & \Rightarrow 10-20 \sin ^2 \cdot \alpha \cos ^2 \alpha=6-5 \cos ^4 \alpha \\ & \Rightarrow 4 \sec ^4 \alpha-20 \tan ^2 \alpha=-5 \\ & \Rightarrow 20 \tan ^2 \alpha-4 \sec ^4 \alpha-5=0 \\ & \Rightarrow 20 \tan ^2 \alpha-4\left(1+\tan ^2 \alpha\right)^2-5=0 \\ & \Rightarrow \tan ^4 \alpha-3 \tan ^2 \alpha+\frac{9}{4}=0 \Rightarrow\left(\tan ^2 \alpha-\frac{3}{2}\right)^2=0 \\ & \Rightarrow \tan ^2 \alpha=\frac{3}{2} \text { then, } \cot ^2 \alpha=\frac{2}{3}\end{aligned}$
Now, $16 \tan ^6 \alpha+27 \cot ^6 \alpha$
$=16 \cdot\left(\frac{3}{2}\right)^3+27\left(\frac{2}{3}\right)^3=16 \times \frac{27}{8}+27 \times \frac{8}{27}=62$