# The Third Algebraic Identity

You learned what $\left(a+b\right)\left(a+b\right)$ and $\left(a-b\right)\left(a-b\right)$ can be expanded to, but what about $\left(a+b\right)\left(a-b\right)$? $\begin{array}{llll}\hfill \left(a+b\right)\left(a-b\right)& ={a}^{2}-ab+ba-{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={a}^{2}-{b}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Rule

### TheThirdAlgebraicIdentity

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

Example 1

Expand $\phantom{\rule{-0.17em}{0ex}}\left(\sqrt{x}+1\right)\phantom{\rule{-0.17em}{0ex}}\left(\sqrt{x}-1\right)$

If you use the third algebraic identity, you get $\begin{array}{llll}\hfill \phantom{\rule{-0.17em}{0ex}}\left(\sqrt{x}+1\right)\phantom{\rule{-0.17em}{0ex}}\left(\sqrt{x}-1\right)& =\phantom{\rule{-0.17em}{0ex}}{\left(\sqrt{x}\right)}^{2}-{1}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =x-1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Sometimes you need to rewrite the expression to be able to use the algebraic identities easily. Look at the following example.

Example 2

Expand $\phantom{\rule{-0.17em}{0ex}}\left(4x-2{x}^{2}\right)\left(x+2\right)$

You can expand this directly by simply multiplying the parentheses together like you did before. But if you want to be a bit more clever about it, you can rewrite the expression inside the first pair of parentheses slightly: $\begin{array}{llll}\hfill \left(4x-2{x}^{2}\right)& =2\cdot 2\cdot x-2\cdot x\cdot x\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\left(2-x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

As $\left(x+2\right)=\left(2+x\right)$, you get