If 2x - y = 2 and xy = $$\frac{3}{2},$$ then what is the value of $${x^3} - \frac{{{y^3}}}{8}?$$

Solution(By Examveda Team)

$$\eqalign{ & 2x - y = 2{\text{ and }}xy = \frac{3}{2} \cr & 2x - y = 2 \cr & {\text{divide '2' both sides}} \cr & x - \frac{y}{2} = 1 \cr & {\text{cube both sides}} \cr & {\left( {x - \frac{y}{2}} \right)^3} = {1^3} \cr & {x^3} - \frac{{{y^3}}}{8} - 3 \times x \times \frac{y}{2}\left( {x - \frac{y}{2}} \right) = {1^3} \cr & {x^3} - \frac{{{y^3}}}{8} = {1^3} + 3 \times \frac{3}{2} \times \frac{1}{2} \times 1 \cr & {x^3} - \frac{{{y^3}}}{8} = 1 + \frac{9}{4} \cr & {x^3} - \frac{{{y^3}}}{8} = \frac{{13}}{4} \cr} $$