If $$x - \frac{3}{x} = 6,\,x \ne 0,$$    then the value of $$\frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}}$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}} \cr & = \frac{{x\left( {{x^3} - \frac{{27}}{{{x^3}}}} \right)}}{{x\left( {x - 3 - \frac{3}{x}} \right)}}.....\left( {\text{i}} \right) \cr & {x^3} - \frac{{27}}{{{x^3}}} \cr & = {6^3} + 3 \times 3 \times 6 \cr & = 216 + 54 \cr & = 270 \cr & \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - \frac{3}{x} - 3}} = \frac{{270}}{{6 - 3}} = 90 \cr} $$