If x > 1 and x^2 + 1x^2 = 83, then the x^3 - 1x^3\, is?

Answer: Option C

$$\begin{array}{rl}& x^2+\frac{1}{x^2}=83\\ & \text{Subtracting 2 from both sides}\\ & \Rightarrow x^2+\frac{1}{x^2}-2=83-2\\ & \Rightarrow x^2+\frac{1}{x^2}-2.x.\frac{1}{x}=83-2\\ & \Rightarrow {\left(x-\frac{1}{x}\right)}^2=81\\ & \Rightarrow x-\frac{1}{x}=9\\ & \text{Take cube on both sides}\\ & \Rightarrow x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=729\\ & \Rightarrow x^3-\frac{1}{x^3}-3\times \text{9}=729\\ & \Rightarrow x^3-\frac{1}{x^3}=729+27\\ & \Rightarrow x^3-\frac{1}{x^3}=756\end{array}$$