Correct Answer: 0

Solution : Given: $(a+\frac{1}{a})^{2}=3$
$\therefore (a+\frac{1}{a})=\sqrt3$
Cubing both sides, we get
$a^3+\frac{1}{a^3}+3×a×\frac{1}{a}(a+\frac{1}{a})=3\sqrt3$
⇒ $a^3+\frac{1}{a^3}+3(\sqrt3)=3\sqrt3$
⇒ $a^3+\frac{1}{a^3}=3\sqrt3-3\sqrt3=0$
To find: $(a^{6}-\frac{1}{a^{6}})$
$(a^{6}-\frac{1}{a^{6}})=(a^{3}-\frac{1}{a^{3}})$$(a^{3}+\frac{1}{a^{3}})$
Putting the value of $a^3+\frac{1}{a^3}$, we get
$(a^{6}-\frac{1}{a^{6}})=(a^{3}-\frac{1}{a^{3}})×0$
⇒ $(a^{6}-\frac{1}{a^{6}})=0$
Hence, the correct answer is 0.