$15{\sin }^4\alpha +10{\cos }^4\alpha =6$

$15{\sin }^4\alpha +10{\cos }^4\alpha =6{\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^2$

${\left(3{\sin }^2\alpha -2{\cos }^2\alpha \right)}^2=0$

$\therefore 3{\sin }^2\alpha -2{\cos }^2\alpha =0$

${\tan }^2\alpha =\frac{2}{3} \& {\cot }^2\alpha =\frac{3}{2}$

$\Rightarrow 27{\sec }^6\alpha +8{cosec}^6\alpha$

$=27{\left({\sec }^2\alpha \right)}^3+8{\left({cosec}^2\alpha \right)}^3$

$=27{\left(1+{\tan }^2\alpha \right)}^3+8{\left(1+{\cot }^2\alpha \right)}^3$

$=27{\left(1+\frac{2}{3}\right)}^3+8{\left(1+\frac{3}{2}\right)}^3$

$=250$