1. Given $x \neq y$, $x^{2}(y+z)=y^{2}(z+x)=2019$.
Then $z^{2}(x+y)-x y z=$ $\qquad$
$$
\begin{array}{l}
\text { II.1.4038. } \\
\text { From } x^{2}(y+z)-y^{2}(z+x) \\
=(x-y)(x y+y z+z x)=0 \text {, }
\end{array}
$$

and given $x \neq y$, we know
$$
\begin{array}{l}
x y+y z+z x=0 \\
\Rightarrow z^{2}(x+y)-y^{2}(z+x) \\
\quad=(z-y)(x y+y z+z x)=0 \\
\Rightarrow z^{2}(x+y)=y^{2}(z+x)=2019 . \\
\text { Also } x y z=-(y z+z x) z \\
=-z^{2}(x+y)=-2019,
\end{array}
$$
$$
\text { Therefore } z^{2}(x+y)-x y z=4038 \text {. }
$$