# How do you factor (a+b)^4 - (a-b)^4?

Jul 14, 2016

=$4 a b \left[{\left(a + b\right)}^{2} + {\left(a - b\right)}^{2}\right]$

#### Explanation:

This is easier than it seems at first glance.
DO NOT multiply out the brackets - that will only make things worse!

This expression is written in the form ${x}^{2} - {y}^{2}$ which is difference of 2 squares.

Let (a+b) be $x$ and (a-b) be $y$, just to make it easier to work with.

${x}^{4} - {y}^{4}$ is also the difference of squares. ${\left({x}^{2}\right)}^{2} = {x}^{4}$

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

=$\textcolor{b l u e}{\left({x}^{2} + {y}^{2}\right)} \textcolor{red}{\left(x + y\right)} \textcolor{\mathmr{and} a n \ge}{\left(x - y\right)} \text{ replace } x \mathmr{and} y$

=$\textcolor{b l u e}{\left[{\left(a + b\right)}^{2} + {\left(a - b\right)}^{2}\right]} \textcolor{red}{\left[\left(a + b\right) + \left(a - b\right)\right]} \textcolor{\mathmr{and} a n \ge}{\left[\left(a + b\right) - \left(a - b\right)\right]}$

$\textcolor{b l u e}{\left[{\left(a + b\right)}^{2} + {\left(a - b\right)}^{2}\right]} \textcolor{red}{\left[\left(2 a\right) \textcolor{\mathmr{and} a n \ge}{\left(2 b\right)}\right]}$
=$4 a b \left[{\left(a + b\right)}^{2} + {\left(a - b\right)}^{2}\right]$