1. Given real numbers $a, b, c$ satisfy $a+b+c=0, a b c>0$, and $x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}, y=a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+$ $c\left(\frac{1}{a}+\frac{1}{b}\right)$. Then the algebraic expression $x^{18}-96 x y+y^{3}=$
$$
=\sqrt{ } .-316
$$

Given $-b c>0$, and $a+b+c=0$, we know that among $a, b, c$, there must be two negative numbers. Without loss of generality, assume $a>0$. Hence $x=-1$.
$$
\begin{array}{l}
\text { ini } \begin{array}{l} 
y=\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b} \\
=\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \\
=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3 . \\
\therefore x^{2 v}-96 x y+y^{3}
\end{array} \\
\therefore=(-1)^{20}-96 \times(-1)(-3)+(-3)^{3} \\
\end{array}
$$