Correct Answer: 9

Solution : Given: $\mathrm{N}+\frac{1}{\mathrm{~N}}=\sqrt{3}$
Squaring both sides, we get:
⇒ $\mathrm{N}^2+2+\frac{1}{\mathrm{~N}^2}=3$
⇒ $\mathrm{N}^2+\frac{1}{\mathrm{~N}^2}=1$
Now, cubing both sides, we get:
⇒ $(\mathrm{N}^2+\frac{1}{\mathrm{~N}^2})^3=1^3$
⇒ $(\mathrm{N}^2)^3 + (\frac{1}{\mathrm{~N}^2})^3 + 3\times\mathrm{N}^2\times\frac{1}{\mathrm{~N}^2}(\mathrm{N}^2+\frac{1}{\mathrm{~N}^2})=1$
⇒ $ \mathrm{N}^6 + \frac{1}{\mathrm{~N}^6} + 3\times1=1$
⇒ $\mathrm{N}^6+\frac{1}{\mathrm{~N}^6}=-2$
Adding 11 on both sides, we get,
$\therefore\mathrm{N}^6+\frac{1}{\mathrm{~N}^6}+11=9$
Hence, the correct answer is 9.