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

$$\eqalign{ & x + \frac{1}{{4x}} = \frac{3}{2} \cr & {\text{Multiply by 2 both sides}} \cr & \therefore 2x + \frac{1}{{2x}} = 3 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2x + \frac{1}{{2x}}} \right)^3} = {\left( 3 \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3.2x.\frac{1}{{2x}}\left( {2x{\text{ + }}\frac{1}{{2x}}} \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3\left( 3 \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} = 18 \cr} $$