1. Let $x=\frac{1}{\sqrt{2}-1}, a$ be the fractional part of $x$, and $b$ be the fractional part of $-x$. Then $a^{3}+b^{3}+3 a b=$ $\qquad$ .
ニ、1. 1 .
Since $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$, and $2<\sqrt{2}+1<3$, therefore,
$$
\begin{array}{l}
a=x-2=\sqrt{2}-1 . \\
\text { Also }-x=-\sqrt{2}-1, \text { and }-3<-\sqrt{2}-1<-2 \text {, so, } \\
b=-x-(-3)=2-\sqrt{2} .
\end{array}
$$

Then $a+b=1$.
Thus $a^{3}+b^{3}+3 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a b$ $=a^{2}-a b+b^{2}+3 a b=(a+b)^{2}=1$.