If $$\left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} {\text{,}}$$     then the value of $$\left( {{x^2} + \frac{1}{{{x^2}}}} \right)$$ $$\left( {x + \frac{1}{x}} \right)$$  is = ?

Solution (By Examveda Team)

$$\eqalign{ & \left( {x - \frac{1}{x}} \right){\text{ = }}\sqrt {21} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2}{\text{ = }}{\left( {\sqrt {21} } \right)^2} = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 21 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 23 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 25 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2}{\text{ = }}{{\text{5}}^2} \cr & \Rightarrow x + \frac{1}{x} = 5 \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {x + \frac{1}{x}} \right) \cr & = 23 \times 5 \cr & = 115 \cr} $$