If ( x + 1/x )^2 = 3 , then the value of (x72 + x66 + x54 + x24 + x6 + 1) is?

Answer: Option A

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