If 6 ^4 + 3 ^4 = 2 , then the value of [ 7 cosec ^6 + 8 ^6 ]^1/3 is?

Answer: Option D

6sin⁴θ + 3cos⁴θ = 2
⇒ 6sin⁴θ + 3(1 - sin²θ )² = 2
⇒ 6sin⁴θ + 3 + 3sin⁴θ - 6sin²θ = 2
⇒ 9sin⁴θ - 6sin²θ + 1 = 0
⇒ (3sin²θ -1)² = 0
⇒ 3sin²θ = 1
⇒ sinθ = $$\frac{1}{\sqrt{3}}$$


Now, $${\left[7{\mathrm{cosec}}^6\theta +8{\sec }^6\theta \right]}^{\frac{1}{3}}$$
$$\begin{array}{rl}& ={\left[7\times {(\sqrt{3})}^6+8{\left(\frac{\sqrt{3}}{\sqrt{2}}\right)}^6\right]}^{\frac{1}{3}}\\ & ={\left[7\times 27+8\times \frac{27}{8}\right]}^{\frac{1}{3}}\\ & ={\left(8\times 27\right)}^{\frac{1}{3}}\\ & =6\end{array}$$