If x^2 + 1x = 41/4, then what is the value of x^3 + 1x^3?

Answer: Option D

$$\begin{array}{rl}& \frac{x^2+1}{x}=4\frac{1}{4}\\ & x+\frac{1}{x}=\frac{17}{4}\\ & \text{On cubing both sides,}\\ & {\left(x+\frac{1}{x}\right)}^3={\left(\frac{17}{4}\right)}^3\\ & x^3+\frac{1}{x^3}+3\times x\times \frac{1}{x}\left(x+\frac{1}{x}\right)=\frac{4913}{64}\\ & x^3+\frac{1}{x^3}+3\left(\frac{17}{4}\right)=\frac{4913}{64}\\ & x^3+\frac{1}{x^3}=\frac{4913}{64}-\frac{51}{4}\\ & x^3+\frac{1}{x^3}=\frac{4097}{64}\end{array}$$