Prove the following. sec^6 x - tan^6 x = 1 + 3 sec^2 x * tan^2 x.

We have,
\[\sec^2 x - \tan^2 x = 1\]
Cubing on both sides, we get
\[\left( \sec^2 x - \tan^2 x \right)^3 = 1^3 \]
\[ \Rightarrow \left( \sec^2 x \right)^3 - \left( \tan^2 x \right)^3 - 3 \times \sec^2 x \times \tan^2 x \times \left( \sec^2 x - \tan^2 x \right) = 1 \left[ \left( a - b \right)^3 = a^3 - b^3 - 3ab\left( a - b \right) \right]\]
\[ \Rightarrow \sec^6 x - \tan^6 x - 3 \sec^2 x \tan^2 x = 1\]
\[ \Rightarrow \sec^6 x - \tan^6 x = 1 + 3 \sec^2 x \tan^2 x\]