If $$x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x}{\text{,}}$$    then the value of $${x^2}{\text{ + }}\frac{1}{{{x^2}}}$$   is?

Solution (By Examveda Team)

$$\eqalign{ & x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x} \cr & \Rightarrow 3x - 2 = \frac{3}{x} \cr & \Rightarrow 3x - \frac{3}{x} = 2 \cr & \Rightarrow 3\left( {x - \frac{1}{x}} \right) = 2 \cr & \Rightarrow x - \frac{1}{x} = \frac{2}{3} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} - 2 = \frac{4}{9} \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = \frac{4}{9} + 2 \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = 2\frac{4}{9} \cr} $$