$$ \begin{aligned} & \text { Given, } 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6 \\\\ & \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2 \\\\ & \Rightarrow \quad 15 \sin ^4 \alpha+10 \cos ^4 \alpha=6\left(\sin ^4 \alpha+\cos ^4 \alpha+2 \sin ^2 \alpha \cos ^2 \alpha\right) \\\\ & \Rightarrow 9 \sin ^4 \alpha+4 \cos ^4 \alpha-12 \sin ^2 \alpha \cos ^2 \alpha=0 \\\\ & \Rightarrow \quad\left(3 \sin ^2 \alpha-2 \cos ^2 \alpha\right)^2=0 \\\\ & \Rightarrow \quad 3 \sin ^2 \alpha-2 \cos ^2 \alpha=0 \\\\ & \Rightarrow \quad 3 \sin ^2 \alpha=2 \cos ^2 \alpha \\\\ & \Rightarrow \quad \tan ^2 \alpha=2 / 3 \\\\ & \therefore \quad \cot ^2 \alpha=3 / 2 \\\\ & \text { Now, } 27 \sec ^6 \alpha+8 \operatorname{cosec}^6 \alpha=27\left(\sec ^2 \alpha\right)^3+8\left(\operatorname{cosec}^2 \alpha\right)^3 \\\\ & =27\left(1+\tan ^2 \alpha\right)^3+8\left(+\cot ^2 \alpha\right)^3 \\\\ & =27\left(1+\frac{2}{3}\right)^3+8\left(1+\frac{3}{2}\right)^3=250 \end{aligned} $$