$6 \cdot 20$ Given $a>1, b$ is a positive rational number, and $a^{b}+a^{-b}=2 \sqrt{3}$, then the value of $a^{b}-$ $a^{-b}$ is 
(A) $-2 \sqrt{3}$.
(B) $2 \sqrt{2}$.
(C) $-2 \sqrt{2}$.
(D) $\pm 2 \sqrt{2}$.
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1985)
[Solution] From $\left(t-\frac{1}{t}\right)^{2}=\left(t+t^{-1}\right)^{2}-4$,
and since $a>1, \quad b$ is a positive rational number, we know
$$
a^{b}-a^{-b}=\sqrt{\left(a^{b}+a^{-b}\right)^{2}-4}=\sqrt{(2 \sqrt{3})^{2}-4}=\sqrt{8}=2 \sqrt{2} \text {. }
$$

Therefore, the answer is $(B)$.