If $${\left( {x + \frac{1}{x}} \right)^2} = 3{\text{,}}$$    then the value of (x⁷² + x⁶⁶ + x⁵⁴ + x²⁴ + x⁶ + 1) is?

Solution(By Examveda Team)

$$\eqalign{ & {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \Rightarrow {x^6} + 1 = 0 \cr & \Rightarrow {x^6} = - 1 \cr & \Rightarrow {x^{72}} + {x^{66}} + {x^{54}} + {x^{24}} + {x^6} + 1 \cr & \Rightarrow {\left( {{x^6}} \right)^{12}} + {\left( {{x^6}} \right)^{11}} + {\left( {{x^6}} \right)^9} + {\left( {{x^6}} \right)^4} + {x^6} + 1 \cr & \Rightarrow {\left( { - 1} \right)^{12}} + {\left( { - 1} \right)^{11}} + {\left( { - 1} \right)^9} + {\left( { - 1} \right)^4} - 1 + 1 \cr & \Rightarrow 1 - 1 - 1 + 1 - 1 + 1 \cr & \Rightarrow 0 \cr} $$