解:
因为$|x|=1,$所以$x=\pm1;$$|y|=\frac{1}{2},$所以$y^{2}=\frac{1}{4}。$
当$x=1$时:
$\begin{aligned}(x^{20})^{3}-x^{3}y^{2}&=1^{60}-1^{3}×\frac{1}{4}\\&=1-\frac{1}{4}\\&=\frac{3}{4}\end{aligned}$
当$x=-1$时:
$\begin{aligned}(x^{20})^{3}-x^{3}y^{2}&=(-1)^{60}-(-1)^{3}×\frac{1}{4}\\&=1+\frac{1}{4}\\&=\frac{5}{4}\end{aligned}$
综上,原式的值为$\frac{3}{4}$或$\frac{5}{4}。$