If 15 ^ 4 α + 10 ^ 4 α = 6, for some α ∈ R, then the value of 27 ^...

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

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

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

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

$\tan^{2}\alpha = \dfrac{2}{3} \text{ and } \cot^{2}\alpha = \dfrac{3}{2}$

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

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

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

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

$= 250$