If $$4{b^2} + \frac{1}{{{b^2}}} = 2,$$ then the value of $$8{b^3} + \frac{1}{{{b^3}}}{\text{ is?}}$$
A. 0
B. 1
C. 2
D. 5
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & 4{b^2} + \frac{1}{{{b^2}}} = 2 \cr & \Rightarrow {\left( {2b} \right)^2} + {\left( {\frac{1}{b}} \right)^2} + 4 - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow 2b + \frac{1}{b} = \sqrt 6 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3 \times 2b \times \frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 6\sqrt 6 - 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 0 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
Join The Discussion