If $${\left( {{\text{ }}a + \frac{1}{a}} \right)^2} = 3{\text{,}}$$    then the value of a¹⁸ + a¹² + a⁶ + 1 is?

Solution (By Examveda Team)

$$\eqalign{ & {\left( {{\text{ }}a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} = - 1 \cr & \Rightarrow \left( {{a^6} + 1} \right) = 0 \cr & {\text{So,}}{a^{18}} + {\text{ }}{a^{12}} + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( {{a^6} + 1} \right) + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$