If $$3x + \frac{1}{{2x}} = 5{\text{,}}$$   then the value of $${\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}}\,{\text{is?}}$$

Solution(By Examveda Team)

$$\eqalign{ & 3x + \frac{1}{{2x}} = 5 \cr & \Rightarrow {\text{Multiply both sides by }}\frac{2}{3} \cr & \therefore 3x \times \frac{2}{3} + \frac{1}{2}x \times \frac{2}{3} = 5 \times \frac{2}{3} \cr & \Rightarrow 2x + \frac{1}{{3x}} = \frac{{10}}{3} \cr & \therefore {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 3.2x.\frac{1}{{3x}}\left( {{\text{ 2}}x{\text{ + }}\frac{1}{{3x}}} \right) = {\left( {\frac{{10}}{3}} \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 2\left( {\frac{{10}}{3}} \right) = \left( {\frac{{1000}}{{27}}} \right) \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000}}{{27}} - \frac{{20}}{3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000 - 180}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{820}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = 30\frac{{10}}{{27}} \cr} $$