If ( x - 1/x ) = 21 , then the value of ( x^2 + 1x^2 ) ( x + 1/x ) is = ?

Answer: Option C

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