If $$x > 1$$  and $${x^2} + \frac{1}{{{x^2}}} = 83,$$   then the $${x^3} - \frac{1}{{{x^3}}}\,{\text{is?}}$$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\text{Subtracting 2 from both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} = 83 - 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & \Rightarrow x - \frac{1}{x} = 9 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3{\text{ }}\left( {x - \frac{1}{x}} \right) = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times {\text{9}} = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 729 + 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$