If $$\frac{4}{3}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 110\frac{2}{3},$$     find $$\frac{1}{9}\left( {{x^3} - \frac{1}{{{x^3}}}} \right),$$   where x > 0.

Solution (By Examveda Team)

$$\eqalign{ & \frac{4}{3}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 110\frac{2}{3} \cr & {x^2} + \frac{1}{{{x^2}}} = \frac{{332}}{3} \times \frac{3}{4} \cr & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\left( {x + \frac{1}{x}} \right)^2} = {x^2} + \frac{1}{{{x^2}}} - 2 \times x \times \frac{1}{x} \cr & {\left( {x + \frac{1}{x}} \right)^2} = 83 - 2 \cr & {\left( {x + \frac{1}{x}} \right)^2} = 81 \cr & x + \frac{1}{x} = 9 \cr & {\text{Hence,}} \cr & \frac{1}{9}\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = \frac{1}{9}\left[ {{{\left( {x - \frac{1}{x}} \right)}^3} + 3 \times \left( {x - \frac{1}{x}} \right)} \right] \cr & = \frac{1}{9}\left[ {729 + 3 \times 9} \right] \cr & = 84 \cr} $$