2.064. $\frac{2 b+a-\frac{4 a^{2}-b^{2}}{a}}{b^{3}+2 a b^{2}-3 a^{2} b} \cdot \frac{a^{3} b-2 a^{2} b^{2}+a b^{3}}{a^{2}-b^{2}}$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}b \neq 0, \\ b \neq-3 a, \\ b \neq \pm a .\end{array}\right.$

$$
\begin{aligned}
& \frac{2 b+a-\frac{4 a^{2}-b^{2}}{a}}{b^{3}+2 a b^{2}-3 a^{2} b} \cdot \frac{a^{3} b-2 a^{2} b^{2}+a b^{3}}{a^{2}-b^{2}}=\frac{\frac{2 a b-a^{2}-4 a^{2}+b^{2}}{a}}{b\left(b^{2}+2 a b-3 a^{2}\right)} \times \\
& \times \frac{a b\left(a^{2}-2 a b+b^{2}\right)}{(a-b)(a+b)}=\frac{\left(a^{2}+2 a b+b^{2}\right)-4 a^{2}}{a b(b+3 a)(b-a)} \cdot \frac{a b(a-b)^{2}}{(a-b)(a+b)}= \\
& =\frac{(a+b)^{2}-4 a^{2}}{-(b+3 a)(a+b)}=\frac{(a+b-2 a)(a+b+2 a)}{-(b+3 a)(a+b)}= \\
& =\frac{(b-a)(b+3 a)}{-(b+3 a)(a+b)}=-\frac{b-a}{a+b}=\frac{a-b}{a+b}
\end{aligned}
$$

Answer: $\frac{a-b}{a+b}$.