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} $.