Prove 2^(1/2) + 6^(1/2) is irrational?

I know how to prove that $\displaystyle \sqrt{2}$ and $\displaystyle \sqrt{6}$ are both themselves irrational, but how would you go about proving that

$\displaystyle \sqrt{2} + \sqrt{6}$ is irrational?

I've tried making the assumption that $\displaystyle \sqrt{2} + \sqrt{6} = \frac{a}{b}$ where $\displaystyle a$ and $\displaystyle b$ are both natural numbers, but it hasn't gotten me anywhere yet. I've manipulated the expression and made some observations when I see certain expressions in the form of $\displaystyle 2n$ or $\displaystyle 2n+1$, therefore being either even or odd, but I can't seem to get to a proper proof. I have though, after making the rational assumption, run into multiple "contradictions", things that just can't be equal. But I'm not certain if these are the type of contradictions that count towards proving the irrationality, or if I made a simple mistake to arrive at the contradictions.

I'm also trying to prove the irrationality of $\displaystyle \sqrt{2} + \sqrt{3}$, its next to the other question in my textbook, and I also have had the same experience with this number as well. thanks in advance for any help.