If x + 1/x = 3 , then the value of x18 + x12 + x6 + 1 is?

Answer: Option A

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