If $${x^2} + \frac{1}{{{x^2}}} = 66{\text{,}}$$    then the value of $$\frac{{{x^2} - 1 + 2x}}{x}$$   = ?

Solution (By Examveda Team)

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 66 \cr & \text{Subtract 2 from both sides} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 66 - 2 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 64 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = {\left( 8 \right)^2} \cr & \Rightarrow x - \frac{1}{x} = \pm 8 \cr & \Rightarrow \frac{{{x^2} - 1 + 2x}}{x} \cr & \Rightarrow \frac{{\frac{{{x^2}}}{x} - \frac{1}{x} + \frac{{2x}}{x}}}{{\frac{x}{x}}} \cr & \Rightarrow \frac{{\left( {x - \frac{1}{x}} \right) + 2}}{1} \cr & {\text{When , }}x - \frac{1}{x} = + 8 \cr & {\text{Then,}} \cr & \left( {x - \frac{1}{x}} \right) + 2 = 8 + 2 = 10 \cr & {\text{When }}x - \frac{1}{x} = - 8 \cr & \Rightarrow x - \frac{1}{x} = - 8 + 2 \cr & \Rightarrow x - \frac{1}{x} = - 6 \cr & \therefore \left( {10, - 6} \right) \cr} $$