若 $a>1, b>0, a^{b}+a^{-b}=2 \sqrt{2}$, 则 $a^{b}-a^{-b}$ 等于( )
选 D $\because a>1, b>0, \therefore a^{b}>a^{-b},\left(a^{b}-a^{-b}\right)^{2}=\left(a^{b}+a^{-b}\right)^{2}-4=(2 \sqrt{2})^{2}-4=4$, $\therefore a^{b}-a^{-b}=2$.