If $$\left( {4{b^2} + \frac{1}{{{b^2}}}} \right){\text{ = 2,}}$$     then $$\left( {8{b^3} + \frac{1}{{{b^3}}}} \right)$$   = ?

Solution (By Examveda Team)

$$\eqalign{ & \left( {4{b^2} + \frac{1}{{{b^2}}}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - {\text{4 = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow \left( {2b + \frac{1}{b}} \right) = \sqrt 6 \cr & \left( {{\text{Cubeing the both sides}}} \right) \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3\times2b\times\frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) = 0 \cr} $$