If $$4{b^2} + \frac{1}{{{b^2}}} = 2,$$    then the value of $$8{b^3} + \frac{1}{{{b^3}}}{\text{ is?}}$$

Solution (By Examveda Team)

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