15. Let the real number $r$ satisfy $r^{3}-r-1=0$, and $r+\sqrt{2}$ be a root of the polynomial $P(x)=0$. If $p(x)$ has a leading coefficient of 1, and $p(x)$ is the polynomial with integer coefficients of the lowest degree that satisfies the condition, then $p(x)=$ $\qquad$ .
$$
\begin{array}{l}
\text { Let } t=r+\sqrt{2} \text {, then }(t-\sqrt{2})^{3}-(t-\sqrt{2})-1 \Rightarrow t^{3}-3 \sqrt{2} t^{2}+5 t-\sqrt{2}-1=0 \\
\Rightarrow\left(t^{3}+5 t-1\right)-\left(3 t^{2}+1\right) \sqrt{2}=0 \Rightarrow\left(t^{3}+5 t-1\right)^{2}-2\left(3 t^{2}+1\right)^{2}=0 \\
\Rightarrow\left(t^{6}+25 t^{2}+1+10 t^{4}-2 t^{3}-10 t\right)-2\left(9 t^{4}+6 t^{2}+1\right)=0 \\
\Rightarrow t^{6}-8 t^{4}-2 t^{3}+13 t^{2}-10 t-1=0 \text {. }
\end{array}
$$

So $p(x)=x^{6}-8 x^{4}-2 x^{3}+13 x^{2}-10 x-1$.