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

##### 1 Answer

Feb 13, 2017

#### Answer:

#### Explanation:

Note that:

#(A+B)(A-B) = A^2-B^2#

So we find:

#(asqrt(b)+csqrt(f))(asqrt(b)-csqrt(f)) = (asqrt(b))^2-(csqrt(f))^2#

#color(white)((asqrt(b)+csqrt(f))(asqrt(b)-csqrt(f))) = a^2b-c^2f#