jmc

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 $$\begin{array}{rl}{x}^2& =(\sqrt{7}+\sqrt{5}{)}^2=7+2\sqrt{35}+5=12+2\sqrt{35},\\ y^2& =(\sqrt{7}-\sqrt{5}{)}^2=7-2\sqrt{35}+5=12-2\sqrt{35}.\end{array}$$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 $$\begin{array}{rl}{x}^6& =x^2\cdot x^4\\ & =(12+2\sqrt{35})(284+48\sqrt{35})\\ & =12\cdot 284+12\cdot 48\sqrt{35}+2\sqrt{35}\cdot 284+2\cdot \sqrt{35}\cdot 48\cdot \sqrt{35}\\ & =6768+1144\sqrt{35}.\end{array}$$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 $\boxed{13535}.$