Twisted Binomial

Let $x=\sqrt{6},y=\sqrt{5}$ .

Thus $(x+y)^6+(x-y)^6=2\left(x^2+y^2\right)\left(x^4+14x^2y^2+y^4\right)=10582.$

But $0<(x-y)^6<1$ , so $\boxed{\left\lfloor\left(\sqrt{6}+\sqrt{5}\right)^6\right\rfloor=10581}$