Question

Write the following cubes in expanded form: (x - $\frac{2y}{3}{)}^3$

Answer 1

(x - $\frac{2y}{3}{)}^3$

Using identity ${(x-y)}^3$ = $x^3$ - $y^3$ - 3xy(x - y)

= $x^3$ - $(\frac{2y}{3}{)}^3$ - 3x$\frac{2y}{3}$(x-$\frac{2y}{3}$)

= $x^3$ - 8$\frac{y^3}{27}$ - 2xy(x-$\frac{2y}{3}$)

= $x^3$ - 8$\frac{y^3}{27}$ - 2$x^2$y + 4x$\frac{y^2}{3}$