jmc

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 $$(a^3+3a^{2}b\sqrt{2}+6a{b}^2+2b^3\sqrt{2})-3(a^2+2ab\sqrt{2}+2b^2)-(a+b\sqrt{2})-\sqrt{2}=0,$$so $$(a^3+6a{b}^2-3a^2-6b^2-a)+(3a^{2}b+2b^3-6ab-b-1)\sqrt{2}=0.$$Hence, $$\begin{array}{rl}{a}^3+6a{b}^2-3a^2-6b^2-a& =0,\\ 3a^{2}b+2b^3-6ab-b-1& =0.\end{array}$$From the first equation, $$6a{b}^2-6b^2=-a^3+3a^2+a,$$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 $x=\boxed{2+\sqrt{2}}.$