jmc

Let $u$ and $v$ be the roots of $x^2-2x-1=0,$ which, by the quadratic formula, are $1\pm \sqrt{2}.$

If $x^6+ax+b=0,$ then by Factor Theorem, $$\begin{array}{rl}{u}^6+au+b& =0,\\ v^6+av+b& =0.\end{array}$$Adding these equations, we get $$a(u+v)+2b+u^6+v^6=0,$$so $2a+2b=-(u^6+v^6).$

Now $$\begin{array}{rl}{u}^6+v^6& =(1+\sqrt{2}{)}^6+(1-\sqrt{2}{)}^6\\ & =1+\left(\genfrac{}{}{0ex}{}{6}{1}\right)\sqrt{2}+\left(\genfrac{}{}{0ex}{}{6}{2}\right)(\sqrt{2}{)}^2+(\genfrac{}{}{0ex}{}{6}{3})(\sqrt{2}{)}^3+\left(\genfrac{}{}{0ex}{}{6}{4}\right)(\sqrt{2}{)}^4+(\genfrac{}{}{0ex}{}{6}{5})(\sqrt{2}{)}^5+(\sqrt{2}{)}^6\\ & +1-\left(\genfrac{}{}{0ex}{}{6}{1}\right)\sqrt{2}+\left(\genfrac{}{}{0ex}{}{6}{2}\right)(\sqrt{2}{)}^2-(\genfrac{}{}{0ex}{}{6}{3})(\sqrt{2}{)}^3+\left(\genfrac{}{}{0ex}{}{6}{4}\right)(\sqrt{2}{)}^4-(\genfrac{}{}{0ex}{}{6}{5})(\sqrt{2}{)}^5+(\sqrt{2}{)}^6\\ & =2(1+15\cdot 2+15\cdot 4+8)\\ & =198,\end{array}$$so $a+b=-198/2=\boxed{-99}.$