Description for Correct answer:
\( \Large a+\frac{1}{a}=\sqrt{3} \)

On squaring both sides

=>\( \Large a^{2}+\frac{1}{a^{2}}+2=\sqrt{3} \)

=>\( \Large a^{2}+\frac{1}{a^{2}}=1 \)

Now, multiplying Eqs. (i) and (ii),

\( \Large \left(a+\frac{1}{a}\right) \left(a^{2}+\frac{1}{a^{2}}\right)=\sqrt{3} \)

=>\( \Large a^{3}+\frac{a}{a^{2}}+\frac{a^{2}}{a}+\frac{1}{a^{3}}=\sqrt{3} \)

=>\( \Large a^{3}+\frac{1}{a^{3}}+ \left(\frac{1}{a}+a\right)=\sqrt{3} \)

=>\( \Large a^{3}+\frac{1}{a^{3}}+\sqrt{3}=\sqrt{3} \) [from Eq. (i)]

=>\( \Large a^{3}+\frac{1}{^{3}}=0 \)

=>\( \Large a^{6}= -1 \)

Therefore, \( \Large a^{6}-\frac{1}{a^{6}}+2 = \left(-1\right)^{6}-\frac{1}{ \left(-1\right)^{6}}+2 \)

= 1-1+2=2