If $${x^2} - 3x + 1 = 0,$$    then the value of $${x^3}{\text{ + }}\frac{1}{{{x^3}}}\,{\text{is?}}$$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & {\text{Divide by }}x \cr & \frac{{{x^2}}}{x} + \frac{1}{x} = \frac{{3x}}{x} \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 18 \cr} $$