# How do you use the binomial theorem to expand (x^(2/3)-y^(1/3))^3?

Jul 27, 2017

The answer is $= {x}^{2} - 3 {x}^{\frac{4}{3}} {y}^{\frac{1}{3}} + 3 {x}^{\frac{2}{3}} {y}^{\frac{2}{3}} - y$

#### Explanation:

We need

${\left(a - b\right)}^{3} = \left(\begin{matrix}3 \\ 0\end{matrix}\right) {a}^{3} - \left(\begin{matrix}3 \\ 1\end{matrix}\right) {a}^{2} b + \left(\begin{matrix}3 \\ 2\end{matrix}\right) a {b}^{2} - \left(\begin{matrix}3 \\ 3\end{matrix}\right) {b}^{3}$

${\left(a - b\right)}^{3} = {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3}$

In our case,

$a = {x}^{\frac{2}{3}}$

and $b = {y}^{\frac{1}{3}}$

Therefore,

$\left({x}^{\frac{2}{3}} - {y}^{\frac{1}{3}}\right) = {\left({x}^{\frac{2}{3}}\right)}^{3} - 3 \cdot {\left({x}^{\frac{2}{3}}\right)}^{2} \cdot {y}^{\frac{1}{3}} + 3 \cdot {x}^{\frac{2}{3}} \cdot {\left({y}^{\frac{1}{3}}\right)}^{2} - {\left({y}^{\frac{1}{3}}\right)}^{3}$

$= {x}^{2} - 3 {x}^{\frac{4}{3}} {y}^{\frac{1}{3}} + 3 {x}^{\frac{2}{3}} {y}^{\frac{2}{3}} - y$