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

142
263
3115
4120

Answer:- 1

Explanation:-

Solution:

\begin{aligned} (x - \frac{1}{x}) = \sqrt{21} \\ \Rightarrow {(x - \frac{1}{x})}^{2} = {(\sqrt{21})}^{2} \\ = 21 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} - 2 = 21 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} = 23 \\ \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 25 \\ \Rightarrow {(x + \frac{1}{x})}^{2} = 5^{2} \\ \Rightarrow x + \frac{1}{x} = 5 \\ ∴ (x^{2} + \frac{1}{x^{2}}) (x + \frac{1}{x}) \\ = 23 \times 5 \\ = 115 \end{aligned}

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