Question:
The polynomial $x^6 + ax + b$ is divisible by $x^2 - 2x - 1.$  Find $a + b.$

Answer:
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{align*}
u^6 + au + b &= 0, \\
v^6 + av + b &= 0.
\end{align*}Adding these equations, we get
\[a(u + v) + 2b + u^6 + v^6 = 0,\]so $2a + 2b = -(u^6 + v^6).$

Now
\begin{align*}
u^6 + v^6 &= (1 + \sqrt{2})^6 + (1 - \sqrt{2})^6 \\
&= 1 + \binom{6}{1} \sqrt{2} + \binom{6}{2} (\sqrt{2})^2 + \binom{6}{3} (\sqrt{2})^3 + \binom{6}{4} (\sqrt{2})^4 + \binom{6}{5} (\sqrt{2})^5 + (\sqrt{2})^6 \\
&\quad + 1 - \binom{6}{1} \sqrt{2} + \binom{6}{2} (\sqrt{2})^2 - \binom{6}{3} (\sqrt{2})^3 + \binom{6}{4} (\sqrt{2})^4 - \binom{6}{5} (\sqrt{2})^5 + (\sqrt{2})^6 \\
&= 2(1 + 15 \cdot 2 + 15 \cdot 4 + 8) \\
&= 198,
\end{align*}so $a + b = -198/2 = \boxed{-99}.$