If $$\left( {4{b^2} + \frac{1}{{{b^2}}}} \right){\text{ = 2,}}$$ then $$\left( {8{b^3} + \frac{1}{{{b^3}}}} \right)$$ = ?
A. 0
B. 1
C. 2
D. 5
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \left( {4{b^2} + \frac{1}{{{b^2}}}} \right){\text{ = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - {\text{4 = 2}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow \left( {2b + \frac{1}{b}} \right) = \sqrt 6 \cr & \left( {{\text{Cubeing the both sides}}} \right) \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3\times2b\times\frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow \left( {8{b^3} + \frac{1}{{{b^3}}}} \right) = 0 \cr} $$Related Questions on Simplification
A. 20
B. 80
C. 100
D. 200
E. None of these
A. Rs. 3500
B. Rs. 3750
C. Rs. 3840
D. Rs. 3900
E. None of these
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