If $$x + \frac{1}{{4x}} = \frac{3}{2}{\text{,}}$$ find the value of $${\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} = ?$$
A. 18
B. 36
C. 24
D. 16
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & x + \frac{1}{{4x}} = \frac{3}{2} \cr & {\text{Multiply by 2 both sides}} \cr & \therefore 2x + \frac{1}{{2x}} = 3 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2x + \frac{1}{{2x}}} \right)^3} = {\left( 3 \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3.2x.\frac{1}{{2x}}\left( {2x{\text{ + }}\frac{1}{{2x}}} \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3\left( 3 \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} = 18 \cr} $$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