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

Solution(By Examveda Team)

$$\eqalign{ & x + \frac{1}{x} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \frac{{{x^3} + \frac{1}{x}}}{{{x^2} - x + 1}}{\text{ }}\left( {{\text{Divide by }}x} \right) \cr & \Rightarrow \frac{{\frac{{{x^3}}}{x} + \frac{1}{{{x^2}}}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x - 1 + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} \cr & \therefore x + \frac{1}{x} = 3 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 9 - 2 = 7 \cr & \therefore \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} = \frac{7}{{3 - 1}} = \frac{7}{2} \cr} $$