TITLE: Greatest integer $n$ where $n \lt (\sqrt5 +\sqrt7)^6$
QUESTION [6 upvotes]: 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!

REPLY [2 votes]: Take the binomial expansion. At $n=6$ pascals is $1$ $6$ $15$ $20$ $15$ $6$ $1$. So $(\sqrt{5}+\sqrt{7})^{6}=$ (grouping $\Bbb N$ seperately from $\Bbb R$) $5^3+7^3+15(5^2 7+7^2 5) + \sqrt{35}(6(5^2+7^2)+20*5*7) = 6768+\sqrt{35}(1144)$. In order to estimate our n, $\sqrt{35}$ must be estimated to 3 places giving us $5.916*1144\approx6767$ rounding down, and $6767+6768=13535$ as the greatest  natural number less than $(\sqrt{5}+\sqrt{7})^{6}$.