解:$\frac {a^3-4ab^2}{a^3-4a^2b+4ab^2}=\frac {a(a^2-4b^2)}{a(a^2-4ab+4b^2)}=\frac {a(a-2b)(a+2b)}{a(a-2b)^2}=\frac {a+2b}{a-2b}$
$ $当$a=2,b=-\frac {1}{2}$时,
$ $原式$=\frac {2+2×(-\frac {1}{2})}{2-2×(-\frac {1}{2})}=\frac {2-1}{2+1}=\frac {1}{3}$