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

Question

1025 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-13T22:38:53

$(asqrt(b)+csqrt(f))(asqrt(b)-csqrt(f)) = a^2b-c^2f$

Note that:

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

So we find:

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

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

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