If 2x - y = 2 and xy = 3/2, then what is the value of x^3 - y^38?

Answer: Option A

$$\begin{array}{rl}& 2x-y=2\text{ and }xy=\frac{3}{2}\\ & 2x-y=2\\ & \text{divide ’2’ both sides}\\ & x-\frac{y}{2}=1\\ & \text{cube both sides}\\ & {\left(x-\frac{y}{2}\right)}^3=1^3\\ & x^3-\frac{y^3}{8}-3\times x\times \frac{y}{2}\left(x-\frac{y}{2}\right)=1^3\\ & x^3-\frac{y^3}{8}=1^3+3\times \frac{3}{2}\times \frac{1}{2}\times 1\\ & x^3-\frac{y^3}{8}=1+\frac{9}{4}\\ & x^3-\frac{y^3}{8}=\frac{13}{4}\end{array}$$