2.217. $\frac{b^{-1 / 6} \cdot \sqrt{a^{3} b} \cdot \sqrt[3]{a^{3} b}-\sqrt{a^{3} b^{2}} \cdot \sqrt[3]{b^{2}}}{\left(2 a^{2}-b^{2}-a b\right) \cdot \sqrt[6]{a^{9} b^{4}}}:\left(\frac{3 a^{3}}{2 a^{2}-a b-b^{2}}-\frac{a b}{a-b}\right)$
## Solution.

Domain of definition: $\left\{\begin{array}{l}b \neq 0, \\ a b>0, \\ a \neq b, \\ a \neq-\frac{b}{.2} .\end{array}\right.$

$$
\begin{aligned}
& \frac{b^{-1 / 6} \cdot \sqrt{a^{3} b} \cdot \sqrt[3]{a^{3} b}-\sqrt{a^{3} b^{2}} \cdot \sqrt[3]{b^{2}}}{\left(2 a^{2}-b^{2}-a b\right) \cdot \sqrt[6]{a^{9} b^{4}}}:\left(\frac{3 a^{3}}{2 a^{2}-a b-b^{2}}-\frac{a b}{a-b}\right)= \\
& =\frac{\sqrt[6]{b^{-1}} \cdot \sqrt[6]{a^{9} b^{3}} \cdot \sqrt[6]{a^{6} b^{2}}-\sqrt[6]{a^{9} b^{6}} \cdot \sqrt[6]{b^{4}}}{(a-b)(2 a+b) \sqrt[6]{a^{9} b^{4}}}:\left(\frac{3 a^{3}}{(a-b)(2 a+b)}-\frac{a b}{a-b}\right)=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\sqrt[6]{a^{15} b^{5}}-\sqrt[6]{a^{9} b^{10}}}{(a-b)(2 a+b) \sqrt[6]{a^{9} b^{4}}}: \frac{3 a^{3}-a b(2 a+b)}{(a-b)(2 a+b)}=\frac{\sqrt[6]{a^{9} b^{4}(a-b)}}{(a-b)(2 a+b) \sqrt[6]{a^{9} b^{4}}}: \\
& : \frac{3 a^{3}-2 a^{2} b-a b^{2}}{(a-b)(2 a+b)}=\frac{1}{2 a+b} \cdot \frac{(a-b) (2 a+b)}{3 a^{3}-2 a^{2} b-a b^{2}}=\frac{a-b}{\left(2 a^{3}-2 a^{2} b\right)+\left(a^{3}-a b^{2}\right)}= \\
& =\frac{a-b}{2 a^{2}(a-b)+a\left(a^{2}-b^{2}\right)}=\frac{a-b}{2 a^{2}(a-b)+a(a-b)(a+b)}= \\
& =\frac{a-b}{(a-b)\left(2 a^{2}+a(a+b)\right)}=\frac{1}{3 a^{2}+a b}=\frac{1}{a(3 a+b)}
\end{aligned}
$$

Answer: $\frac{1}{a(3 a+b)}$.