If $$x + \frac{1}{x} = 3{\text{,}}$$   then the value of $$\left( {{x^5} + \frac{1}{{{x^5}}}} \right)\,{\text{is?}}$$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \left( {{\text{Squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{On cubing both sides}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 18 \times 7 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + \left( {x + \frac{1}{x}} \right) = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + 3 = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) = 123 \cr} $$