If ( 4b^2 + 1b^2 ) = 2, then ( 8b^3 + 1b^3 ) = ?

Answer: Option A

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