Find the positive root of
\[x^3 - 3x^2 - x - \sqrt{2} = 0.\]
Since we have a coefficient of $\sqrt{2},$ we can guess that the positive root is of the form $a + b \sqrt{2},$ where $a$ and $b$ are integers.  So, let $x = a + b \sqrt{2}.$  Substituting, we get
\[(a + b \sqrt{2})^3 - 3(a + b \sqrt{2})^2 - (a + b \sqrt{2}) - \sqrt{2} = 0.\]This expands as
\[(a^3 + 3a^2 b \sqrt{2} + 6ab^2 + 2b^3 \sqrt{2}) - 3(a^2 + 2ab \sqrt{2} + 2b^2) - (a + b \sqrt{2}) - \sqrt{2} = 0,\]so
\[(a^3 + 6ab^2 - 3a^2 - 6b^2 - a) + (3a^2 b + 2b^3 - 6ab - b - 1) \sqrt{2} = 0.\]Hence,
\begin{align*}
a^3 + 6ab^2 - 3a^2 - 6b^2 - a &= 0, \\
3a^2 b + 2b^3 - 6ab - b - 1 &= 0.
\end{align*}From the first equation,
\[6ab^2 - 6b^2 = -a^3 + 3a^2 + a,\]so
\[6b^2 (a - 1) = -(a^3 - 3a^2 - a).\]Thus, $a - 1$ divides $a^3 - 3a^2 - a.$  Since $a - 1$ divides $(a - 1)(a - 3)(a + 1) = a^3 - 3a^2 - a + 3,$ $a - 1$ divides 3.  This means $a - 1$ can be $-3,$ $-1,$ 1, or 3, so $a$ is $-2$, 0, 2, or 4.

If $a = -2,$ then $b^2 = -1,$ which has no solutions.

If $a = 0,$ then $b^2 = 0,$ so $b = 0,$ which does not work.

If $a = 2,$ then $b^2 = 1,$ so $b = -1$ or $b = 1.$  Only $a = 2$ and $b = 1$ satisfy the second equation.

If $a = 4,$ then $b^2 = -\frac{2}{3},$ which has no solutions.

Therefore, $a = 2$ and $b = 1$ works, so $x = \boxed{2 + \sqrt{2}}.$