2. Given $a<b<0$, and $\frac{a}{b}+\frac{b}{a}=6$. Then $\left(\frac{a+b}{a-b}\right)^{3}=$ $\qquad$
2. $2 \sqrt{2}$.
$$
\begin{array}{l}
\text { Given } \frac{a}{b}+\frac{b}{a}=6 \Rightarrow a^{2}+b^{2}=6 a b \\
\Rightarrow(a+b)^{2}=8 a b,(a-b)^{2}=4 a b .
\end{array}
$$

Since $a<b<0$, we have,
$$
\begin{array}{l}
a+b=-2 \sqrt{2 a b}, \\
a-b=-2 \sqrt{a b} . \\
\text { Therefore }\left(\frac{a+b}{a-b}\right)^{3}=(\sqrt{2})^{3}=2 \sqrt{2} .
\end{array}
$$