Example 1 When $a^{3}-a-1=0$, $a+\sqrt{2}$ is a root of some integer-coefficient polynomial. Find the integer-coefficient polynomial with the lowest degree and the leading coefficient of 1 that satisfies the given condition.
(1997 Japan Mathematical Olympiad Preliminary Exam)
Let $x=a+\sqrt{2}$, then $0=a^{3}-a-1=(x-\sqrt{2})^{2}-(x-\sqrt{2})-1$ $=\left(x^{3}+5 x+1\right)+\sqrt{2}\left(-3 x^{2}-1\right)$.

From this, we get $x^{3}+5 x-1=\sqrt{2}\left(3 x^{2}+1\right)$.
Squaring both sides, we have $x^{6}+10 x^{4}-2 x^{3}+25 x^{2}-10 x+1=18 x^{4}+12 x^{2}+2$, rearranging and simplifying, we get $x^{6}-8 x^{4}-2 x^{3}+13 x^{2}-10 x-1=0$.
Therefore, the required polynomial is $x^{6}-8 x^{4}-2 x^{3}+13 x^{2}-10 x-1$.