1. If integers $a, b$ satisfy $\frac{2 \sqrt[3]{4}-b}{\sqrt[3]{4}-a \sqrt[3]{2}+1}$ $=\sqrt[3]{4}+1$, then $a=$ $\qquad$ ,$b=$ $\qquad$
$$
\begin{array}{l}
\text { II. 1. } a=2, b=3 . \\
\text { Let } \sqrt[3]{2}=x \text {, then } \\
\frac{2 x^{2}-b}{x^{2}-a x+1}=x^{2}+1,
\end{array}
$$

which means
$$
\begin{array}{l}
x^{4}-a x^{3}-a x+b+1=0 . \\
\because x^{3}=2, \\
\therefore(2-a) x=2 a-b-1 .
\end{array}
$$
$\because x$ is an irrational number, and $2 a-b-1$ is an integer,
$$
\therefore\left\{\begin{array}{l}
2-a=0, \\
2 a-b-1=0 .
\end{array}\right.
$$

Solving, we get $a=2, b=3$.