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!

We want a lower bound on
$\sqrt{1-x}$.
From $\sqrt{1-x} = (1-x)^{1/2}
=1-x/2-x^2/8-x^3/16 -5x^4/128 ...
$
I will try
$1-x/2-x^2/4$.
$\begin{align}
(1-x/2-x^2/4)^2
&=1-x-x^2(1/4+1/2)+x^3/4+x^4/16\\
&=1-x-3x^2/4+x^3/4+x^4/16\\
&= 1-x-x^2(3/4-x/4-x^2/16)\\
&< 1-x\\
\end{align}
$
for $0 < x < 1/2$,
so $\sqrt{1-x} > 1-x/2-x^2/4$
for $0 < x < 1/2$.
$\sqrt{35} = \sqrt{36-1}
=
6\sqrt{1-1/36}
> 6(1-1/12-1/(4*36))
=6-1/2-1/24
$
so
$6+\sqrt{35} > 12-1/2-1/24
> 12-1$
as you wanted.
