Greatest integer $n$ where $n \lt (\sqrt5 +\sqrt7)^6$ I'm really not sure how to do this. I factored out a power of $3$ and squared so that I have $2^3 (6+\sqrt{35})^3 \gt n$ , and I know that if I can prove that $12^3-1 \le (6+\sqrt{35})^3 \lt 12^3$ I am basically done, but I don't know how to do that. Any help is appreciated. Thanks!

First, a comment on technique. Heuristically, what you're doing is "wrong" for a competition problem in the sense that you're ignoring and throwing away a lot of the structure. (You have $\sqrt{n+1}-\sqrt{n-1}$ raised to a power. This is a contest problem. That's not a coincidence.) Your way might be fine to find some weak bounds to in order to check your final answer, but before I even found the solution myself, I could tell it probably wasn't going to work. 
Whenever you see a problem involving sums of radicals, you should immediately think about taking the conjugate and somehow playing tricks with it. This idea quickly leads to a solution here. The conjugate is $(\sqrt 7 - \sqrt 5)^6$, and it's not hard to notice this is a number less than one. Then 
$$(\sqrt 7 + \sqrt 5)^6 + (\sqrt 7 - \sqrt 5)^6$$
is an integer, and the rest of the solution is clear. 
