If $$x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x},$$    then the value of $${x^3} - \frac{1}{{{x^3}}}$$  is equal to:

$$\eqalign{ & x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x} \cr & 3x - 2 = \frac{3}{x} \cr & 3\left( {x - \frac{1}{x}} \right) = 2 \cr & x - \frac{1}{x} = \frac{2}{3} \cr & {x^3} - \frac{1}{{{x^3}}} = {\left( {\frac{2}{3}} \right)^3} + 3 \times \frac{2}{3} \cr & {x^3} - \frac{1}{{{x^3}}} = \frac{8}{{27}} + 2 \cr & {x^3} - \frac{1}{{{x^3}}} = \frac{{62}}{{27}} \cr} $$