e.$(\frac{x}{2}$ + $\frac{y}{3})^3$ = $(\frac{x}{2})^3$ + $3.(\frac{x}{2})^2\frac{y}{3}$ + $3.\frac{x}{2}(\frac{y}{3})^2$ + $(\frac{y}{3})^3$
=$\frac{x^3}{8}$ + $3.\frac{x^2}{4}.\frac{y}{3}$ + $3.\frac{x}{2}.\frac{y^2}{9}$ + $\frac{y^3}{27}$
=$\frac{x^3}{8}$ + $\frac{x^2y}{4}$ + $\frac{xy^2}{6}$ + $\frac{y^3}{27}$
f. $(\frac{2x}{3}$ - 2y)³ = $(\frac{2x}{3})^3$ - $3.(\frac{2x}{3})^22y$ + $3.\frac{x}{2}(2y)^2$ - $(2y)^3$
=$\frac{8x}{27}$ - $3.\frac{4x}{9}.2y$ + $3.\frac{x}{2}4y^2$ - $8y^3$
= $\frac{8x}{27}$ - $\frac{8xy}{3}$ + $6xy^2$ - $8y^3$
g) (x+y)³ - (x-y)³ = [(x+y)-(x-y)][(x+y)² + (x-y)(x+y) + (x-y)²]
= (x+y-x+y)(x²+2xy +y² + x² - y² + x² - 2xy + y²)
=2y(3x² +y²) = 6x²y + 2y³
