A student asks:
How could I simplify a sum like $(\sqrt 3+\sqrt 2)(\sqrt 3-\sqrt 2)$?
The trick is to treat it like it's an algebraic bracket, like this:
$(x + y)(x - y) = x^2 + yx - xy - y^2$
But then you've got $+yx -xy$ in the middle there, which adds up to nothing:
$(x+y)(x-y) = x^2 - y^2$ (that's the difference of two squares - it comes up a lot).
If you do that with $(\sqrt 3 + \sqrt 2)(\sqrt 3 - \sqrt 2)$, you get $(\sqrt 3)(\sqrt 3) - (\sqrt 2)(\sqrt 2)$.
Now for the clever bit: $\sqrt 3$ is the answer to 'what, multiplied by itself, makes 3?' - so when you times it by itself, you get three! Same goes for $\sqrt 2$ - you end up with
$(\sqrt 3 + \sqrt 2)(\sqrt 3 - \sqrt 2) = 3 - 2 = 1$.
Check it on the calculator if you don't believe me!
Hope that helps!