Answer: $\mathbf B$
Solution:
Given:$$\sin^6\frac{\pi}{81}+\cos^6\frac{\pi}{81} -1 + 3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81} \tag{i}$$
Now,
$\sin^6\frac{\pi}{81} = \left(\sin^2\frac{\pi}{81}\right)^3,\; \text{and}\;\cos^6\frac{\pi}{81} = \left (\cos^2\frac{\pi}{81}\right)^3 \tag{ii}$
Using identity: $a^3 + b^3 = (a+b)^3 -3ab(a+b)\tag{iii}$
and, $\sin^2\theta + \cos^2\theta = 1\tag{iv}$
Using equations $\;\mathrm{(ii)\;, (iii)\;and\;(iv)}$, we get:
$\sin^6\frac{\pi}{81}+\cos^6\frac{\pi}{81}=\left(\sin^2\frac{\pi}{81}\right)^3 + \left(\cos^2\frac{\pi}{81}\right)^3$
$ = \underbrace{\sin^2\frac{\pi}{81} + \cos^2\frac{\pi}{81}}_\text{=1} -3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81}\left(\underbrace{\sin^2\frac{\pi}{81} + \cos^2\frac{\pi}{81}}_\text{=1}\right) $
$= 1 - 3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81} \tag{v}$
From equations $\mathrm {(i)\;and\;(v)}$, we get:
$ \require{cancel} = \cancel {1} - \cancel{3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81}} -\cancel{1} + \cancel{3\sin^2\frac{\pi}{81}.\cos^2\frac{\pi}{81}}\\ = \mathbf 0$
$\therefore \mathbf B$ is the correct answer.
