If 4b^2 + 1b^2 = 2, then the value of 8b^3 + 1b^3 is?

Answer: Option A

$$\begin{array}{rl}& 4b^2+\frac{1}{b^2}=2\\ & \Rightarrow {\left(2b\right)}^2+{\left(\frac{1}{b}\right)}^2+4-4=2\\ & \Rightarrow {\left(2b+\frac{1}{b}\right)}^2-4=2\\ & \Rightarrow {\left(2b+\frac{1}{b}\right)}^2=6\\ & \Rightarrow 2b+\frac{1}{b}=\sqrt{6}\\ & \text{Take cube both sides}\\ & \Rightarrow {\left(2b+\frac{1}{b}\right)}^3={\left(\sqrt{6}\right)}^3\\ & \Rightarrow 8b^3+\frac{1}{b^3}+3\times 2b\times \frac{1}{b}\left(2b+\frac{1}{b}\right)=6\sqrt{6}\\ & \Rightarrow 8b^3+\frac{1}{b^3}+6\sqrt{6}=6\sqrt{6}\\ & \Rightarrow 8b^3+\frac{1}{b^3}=6\sqrt{6}-6\sqrt{6}\\ & \Rightarrow 8b^3+\frac{1}{b^3}=0\end{array}$$