若 $a<0,-1<b<0$, 则有 $(\quad)$
$\because a<0,-1<b<0$,

$\therefore a b>0, a b^{2}<0$.

$\therefore a b>a, a b>a b^{2}$.

$\because a-a b^{2}=a\left(1-b^{2}\right)=a(1+b)(1-b)<0$,

$\therefore a<a b^{2} . \therefore a<a b^{2}<a b$.]