If $$x + \frac{1}{x} = \sqrt 3 {\text{,}}$$   then the value of x¹⁸ + x¹² + x⁶ + 1 is?

$$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & \left( {{\text{Take cube on both sides}}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \therefore {x^6} = - 1 \cr & \therefore {x^{18}} + {x^{12}} + {x^6} + 1 \cr & = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 \cr & = - 1 + 1 - 1 + 1 \cr & = 0 \cr} $$