Prove that $\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta = 1$.

(N/A) We know the fundamental trigonometric identity: $\sin^{2} \theta + \cos^{2} \theta = 1$.
Taking the cube of both sides,we get:
$(\sin^{2} \theta + \cos^{2} \theta)^{3} = (1)^{3}$
Using the algebraic identity $(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)$,where $a = \sin^{2} \theta$ and $b = \cos^{2} \theta$:
$(\sin^{2} \theta)^{3} + (\cos^{2} \theta)^{3} + 3(\sin^{2} \theta)(\cos^{2} \theta)(\sin^{2} \theta + \cos^{2} \theta) = 1$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,the equation becomes:
$\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta (1) = 1$
Thus,$\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta = 1$. Hence proved.