# How do you find the product (n-p)^2(n+p)?

Jun 23, 2017

${n}^{3} - n {p}^{2} - {n}^{2} p + {p}^{3}$

#### Explanation:

We can write ${\left(n - p\right)}^{2}$ as $\left(n - p\right) \left(n - p\right)$:

${\left(n - p\right)}^{2} \left(n + p\right) = \left(n - p\right) \left(n - p\right) \left(n + p\right)$

We can group $\left(n - p\right)$ and $\left(n + p\right)$ and multiply those, since they will result in a difference of squares, which is only two terms instead of three. That will make our future distribution easier.

$= \left(n - p\right) \left\{\left(n - p\right) \left(n + p\right)\right\} = \left(n - p\right) \left({n}^{2} - {p}^{2}\right)$

Now distribute:

$= n \left({n}^{2} - {p}^{2}\right) - p \left({n}^{2} - {p}^{2}\right)$

$= {n}^{3} - n {p}^{2} - {n}^{2} p + {p}^{3}$