# How do you show that if you multiply asqrtb+csqrtf and asqrtb-csqrtf, the product has no radicals?

Feb 13, 2017

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

#### Explanation:

Note that:

$\left(A + B\right) \left(A - B\right) = {A}^{2} - {B}^{2}$

So we find:

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

$\textcolor{w h i t e}{\left(a \sqrt{b} + c \sqrt{f}\right) \left(a \sqrt{b} - c \sqrt{f}\right)} = {a}^{2} b - {c}^{2} f$