$6 \cdot 19$ If $a>1$, and $b$ is a positive rational number, and $a^{b}+a^{-b}=2 \sqrt{2}$, then $a^{b}-a^{-b}$ equals
(A) $\sqrt{6}$.
(B) 2 or -2 .
(C) -2 .
(D) 2 .
(Tianjin, China Junior High School Mathematics Competition, 1983)
[Solution] Consider $\left(a^{b}-a^{-b}\right)^{2}=\left(a^{b}+a^{-b}\right)^{2}-4$, i.e., $\left(a^{b}-a^{-b}\right)^{2}=(2 \sqrt{2})^{2}-4=4$,

Also, $a>1, b>0$, then $a^{b}>a^{-b}$, thus $a^{b}-a^{-b}=2$. Therefore, the answer is $(D)$.