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

Solution (By Examveda Team)

$$\eqalign{ & \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \frac{3}{2}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{x\left\{ {\left( {x + \frac{1}{x}} \right) - 1} \right\}}}{{x\left\{ {\left( {x + \frac{1}{x}} \right) + 1} \right\}}} = \frac{3}{2} \cr & \Rightarrow \frac{{\left( {x + \frac{1}{x}} \right) - 1}}{{\left( {x + \frac{1}{x}} \right) + 1}} = \frac{3}{2} \cr & \,\,\,\,\,\,\,\,\,\,{\text{Let }}\left( {x + \frac{1}{x} = y} \right) \cr & \Rightarrow \frac{{y - 1}}{{y + 1}} = \frac{3}{2} \cr & \Rightarrow 2\left( {y - 1} \right) = 3\left( {y + 1} \right) \cr & \Rightarrow 2y - 2 = 3y + 3 \cr & \Rightarrow y = - 2 - 3 \cr & \Rightarrow y = - 5 \cr & \therefore x + \frac{1}{x} = - 5 \cr & {\text{ }} \cr} $$