So I was doing some problems with a lot of conjugates, and I suddenly remembered that conjugates to the same power will have same terms - i.e., $(\sqrt{3} + \sqrt{2})^6 = 485 + 198\sqrt{6}$, and $(\sqrt{3} - \sqrt{2})^6 = 485 - 198\sqrt{6}$. And it doesn't seem to matter whether the power is even or odd. But how does this happen? Is there a proof about this? (The thing is, I could google this, I think, but I don't know what to type in etc.)
Edit note: I'm sorry for the confusion - I got mixed up with another problem! Now it says to the 6th power.
I think the binomial theorem is the way to think here:
$$(a+\sqrt{b})^n = {n \choose 0}a^n+{n\choose 1}a^{n-1}\sqrt{b} + {n \choose 2}a^{n-2}b +{n \choose 3}a^{n-3}b\sqrt{b}+\cdots$$
and
$$(a-\sqrt{b})^n = {n \choose 0}a^n-{n\choose 1}a^{n-1}\sqrt{b} + {n \choose 2}a^{n-2}b -{n \choose 3}a^{n-3}b\sqrt{b}+\cdots.$$
Note two things:
When the exponent on $\sqrt{b}$ is even, the corresponding terms in both sums are the same. When the exponent is odd, the terms are the same but for the opposite sign.
When the exponent is even, no radical survives and when the exponent is odd there's a $\sqrt{b}$ in the term. If you gather up the terms with and without $\sqrt{b}$, then you end up with something of the form $A+C\sqrt{b}$, for the first sum, and $A-C\sqrt{b}$ for the second.
A similar argument works if you replace $a$ by $\sqrt{a}$.