If ( a + 1/a )^2 = 3 , then find the value of a^30 + a^24 + a^18 + a^12 + a^6 + 1 = ?

Answer: Option A

$$\begin{array}{rl}& {\left(a+\frac{1}{a}\right)}^2=3\\ & \Rightarrow a+\frac{1}{a}=\sqrt{3}\\ & \text{Cube both sides}\\ & \Rightarrow a^3+\frac{1}{a^3}+3\times a\times \frac{1}{a}\left(a+\frac{1}{a}\right)={\left(\sqrt{3}\right)}^3\\ & \Rightarrow a^3+\frac{1}{a^3}+3\left(\sqrt{3}\right)=3\sqrt{3}\\ & \Rightarrow a^3+\frac{1}{a^3}=3\sqrt{3}-3\sqrt{3}\\ & \Rightarrow a^3+\frac{1}{a^3}=0\\ & \Rightarrow a^6+1=0\\ & \therefore a^{30}+a^{24}+a^{18}+a^{12}+a^6+1=?\\ & \Rightarrow a^{24}\left(a^6+1\right)+a^{12}\left(a^6+1\right)+a^6+1\\ & \Rightarrow a^{24}\left(0\right)+a^{12}\left(0\right)+0\\ & \Rightarrow 0\end{array}$$