Let $x = \sqrt{7} + \sqrt{5}$ and $y = \sqrt{7} - \sqrt{5}.$

First, we can square $x = \sqrt{7} + \sqrt{5}$ and $y = \sqrt{7} - \sqrt{5},$ to get
\begin{align*}
x^2 &= (\sqrt{7} + \sqrt{5})^2 = 7 + 2 \sqrt{35} + 5 = 12 + 2 \sqrt{35}, \\
y^2 &= (\sqrt{7} - \sqrt{5})^2 = 7 - 2 \sqrt{35} + 5 = 12 - 2 \sqrt{35}.
\end{align*}Note that $x^2$ and $y^2$ are radical conjugates.  Also, $x^2 y^2 = (12 + 2 \sqrt{35})(12 - 2 \sqrt{35}) = 12^2 - 2^2 \cdot 35 = 4,$ so
\[y^2 = \frac{4}{x^2} = \frac{4}{12 + 2 \sqrt{35}} < 1.\]Then
\[x^4 = (12 + 2 \sqrt{35})^2 = 12^2 + 2 \cdot 12 \cdot 2 \sqrt{35} + 2^2 \cdot 35 = 284 + 48 \sqrt{35},\]and
\begin{align*}
x^6 &= x^2 \cdot x^4 \\
&= (12 + 2 \sqrt{35})(284 + 48 \sqrt{35}) \\
&= 12 \cdot 284 + 12 \cdot 48 \sqrt{35} + 2 \sqrt{35} \cdot 284 + 2 \cdot \sqrt{35} \cdot 48 \cdot \sqrt{35} \\
&= 6768 + 1144 \sqrt{35}.
\end{align*}Then $y^6$ is the radical conjugate of $x^6,$ so $y^6 = 6768 - 1144 \sqrt{35}.$  Hence,
\[x^6 + y^6 = (6768 + 1144 \sqrt{35}) + (6768 - 1144 \sqrt{35}) = 13536.\]Since $0 < y^6 < 1,$ the greatest integer less than $x^6$ is $\boxed{13535}.$
Question:
Find the greatest integer less than $(\sqrt{7} + \sqrt{5})^6.$  (Do not use a calculator!)
Answer:
Let's think step by step.

user
13535
Find the greatest integer less than $(\sqrt{7} + \sqrt{5})^6.$  (Do not use a calculator!)
test