1. When $a^{3}-a-1=0$, $a+\sqrt{2}$ is a root of some polynomial with integer coefficients. Find the polynomial of the lowest degree with leading coefficient 1 that satisfies the above condition.
1. Let $T=a+\sqrt{2}$, then $a=T-\sqrt{2}$, substituting into the original equation, we get $0=a^{3}-a-1=(T-\sqrt{2})^{3}-(T-\sqrt{2})-1$, which is $T^{3}-3 \sqrt{2} T^{2}+5 T-(\sqrt{2}+1)=0$.
Thus, $T^{3}+5 T-1=\sqrt{2}\left(3 T^{2}+1\right)$.
Squaring both sides, we get $T^{6}+25 T^{2}+1+10 T^{4}-2 T^{3}-10 T=2\left(9 T^{4}+1+6 T^{2}\right)$.
Therefore, the required polynomial is $f(x)=x^{6}-8 x^{4}-2 x^{3}+13 x^{2}-10 x-1$.