Question

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

Answer 1

Expand ${\left(x-\frac{2}{3}y\right)}^3$:

Use the formula ${\left(u-v\right)}^3=u^3-v^3-3u^{2}v+3u{v}^2$.

Put $u=x,v=\frac{2}{3}y$.

$\begin{array}{rcl}{\left(x-\frac{2}{3}y\right)}^3& =& x^3-{\left(\frac{2}{3}y\right)}^3-3x^2\left(\frac{2}{3}y\right)+3x{\left(\frac{2}{3}y\right)}^2\\ \therefore {\left(x-\frac{2}{3}y\right)}^3& =& x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2\end{array}$

Hence, expanded form of ${\left(x-\begin{array}{r}\frac{2}{3}y\end{array}\right)}^3$ is $x^3-\frac{8}{27}{y}^3-2x^{2}y+\frac{4}{3}x{y}^2$.