# How do you simplify  (x - y)^3?

Jul 14, 2017

${x}^{3} - 3 {x}^{2} y + 3 x {y}^{2} - {y}^{3}$

#### Explanation:

${\left(x - y\right)}^{3}$

Solution

Well you can use many methods to simplify like:

Using Pascal Triangle which give be $1 , 3 , 3 , 1$ as the expansion..

You can simplify ${\left(x - y\right)}^{3}$ to either $\left(x - y\right) \left(x - y\right) \left(x - y\right) \mathmr{and} {\left(x - y\right)}^{2} \left(x - y\right)$

But using those two will result in same answer which will be in this format $\to$ $1 , 3 , 3 , 1$

Hence $\Rightarrow$ ${\left(x - y\right)}^{3} = \left(x - y\right) \left(x - y\right) \left(x - y\right)$

$\left(x - y\right) \left(x - y\right) \left(x - y\right)$

$\left(x - y\right) \left[\left(x - y\right) \left(x - y\right)\right]$

$\left(x - y\right) \left[{x}^{2} - x y - x y + {y}^{2}\right]$

This lead to the point using difference of two cubes as:

$\left(x - y\right) \left[{x}^{2} - 2 x y + {y}^{2}\right]$

$x \left[{x}^{2} - 2 x y + {y}^{2}\right] - y \left[{x}^{2} - 2 x y + {y}^{2}\right]$

${x}^{3} - 2 {x}^{2} y + x {y}^{2} - {x}^{2} y + 2 x {y}^{2} - {y}^{3}$

Collect like terms

$\textcolor{red}{{x}^{3} - {y}^{3}} \textcolor{b l u e}{- 2 {x}^{2} y - {x}^{2} y} \textcolor{g r e e n}{+ x {y}^{2} + 2 x {y}^{2}}$

$\therefore$ ${x}^{3} - {y}^{3} - 3 {x}^{2} y + 3 x {y}^{2}$ $\to A n s w e r$

If it is the cube of a binomial, it will be in this format $\Rightarrow$ $\textcolor{red}{1} {x}^{3} - \textcolor{red}{3} {x}^{2} y + \textcolor{red}{3} x {y}^{2} - \textcolor{red}{1} {y}^{3}$