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The value of $\sin^6 \frac{\pi}{81} + \cos^6 \frac{\pi}{81}-1+3 \sin ^2 \frac{\pi}{81} \cos^2 \frac{\pi}{81}$  is

  1. $\tan ^6 \frac{\pi}{81}$
  2. $0$
  3. $-1$
  4. None of these
in Numerical Ability
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$\underline{\textbf{Answer: B}\Rightarrow}$

$\underline{\textbf{Solution:}\Rightarrow}$

The above equation can be converted in the form of identity of the form:

$(a+b)^3 = a^3 + b^3 + 3ab(a+b)$

Let $a = \sin^2\frac{\pi}{81}, b = \cos^2\frac{\pi}{81}$

Now,

$=(\sin^2\frac{\pi}{81}+\cos^2\frac{\pi}{81})^3 = (\sin^2\frac{\pi}{81})^3 + (\sin^2\frac{\pi}{81})^3 + 3\sin^2\frac{\pi}{81}cos^2\frac{\pi}{81}\underbrace{(\sin^2\frac{\pi}{81}+\cos^2\frac{\pi}{81})}_\text{=1} = 1$

But the question has extra $-1$. So, answer = $1-1$ = 0

$\therefore \mathbf B$ is the correct option.

edited by

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