If $$\left( {x + \frac{1}{x}} \right){\text{ = }}\sqrt {13} {\text{,}}$$     then the value of $$\left( {{x^3} - \frac{1}{{{x^3}}}} \right)$$   is = ?

$$\eqalign{ & \left( {x + \frac{1}{x}} \right) = \sqrt {13} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} - 4 = {\left( {\sqrt {13} } \right)^2} - 4 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow \left( {x - \frac{1}{x}} \right) = 3 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {3^3} = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times 3 = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 = 36 \cr} $$