# How do you factor 3u^3w^4- 3u^3?

Mar 23, 2016

$3 {u}^{3} {w}^{4} - 3 {u}^{3} = 3 {u}^{3} \left({w}^{2} + 1\right) \left(w + 1\right) \left(w - 1\right)$

#### Explanation:

$3 {u}^{3} {w}^{4} - 3 {u}^{3} = 3 {u}^{3} \left({w}^{4} - 1\right) = 3 {u}^{3} \left[{\left({w}^{2}\right)}^{2} - 1\right] =$

Now you can factor $\left[{\left({w}^{2}\right)}^{2} - 1\right]$ with the remarkable identity:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$ where $a = {w}^{2}$ and $b = 1$

$= 3 {u}^{3} \left[\left({w}^{2} + 1\right) \left({w}^{2} - 1\right)\right] =$

Now $\left({w}^{2} - 1\right)$ can more be factorizide with the same identity

$= 3 {u}^{3} \left[\left({w}^{2} + 1\right) \left(w + 1\right) \left(w - 1\right)\right]$