Supposedly solution is $\frac{a+1}{ab}$, but both problem and solution could be erroneously defined because a book I retyped problem from has a few misprints here and there.
There's no need to post my attempts to solve and unnecessary clutter post.
I just need starting hint. Thanks in advance!
On
First simply this term $$\frac{a^2-ab}{a^2b+b^3}-\frac{2a^2}{b^3-ab^2+a^2b-a^3}$$
$$\frac{a^2-ab}{a^2b+b^3}-\frac{2a^2}{b^3-ab^2+a^2b-a^3}=\frac{a^2-ab}{b(a^2+b^2)}-\frac{2a^2}{(b-a)(a^2+b^2)}=\frac{(b-a)(a^2-ab)-2a^2b}{b(b-a)(a^2+b^2)}=\frac{-a^3-ab^2}{b(b-a)(a^2+b^2)}=\frac{-a}{b^2-ab}$$
Now
$$(\frac{-a}{b^2-ab})\cdot(1-\frac{b-1}{a}-\frac{b}{a^2})=(\frac{-a}{b^2-ab})\cdot(\frac{a^2-ab+a-b}{b^2})=\frac{(b-a)(a+1)}{ab(b-a)}=\frac{a+1}{ab}$$
$$\left(\frac{a^2-ab}{a^2b+b^3}-\frac{2a^2}{b^3-ab^2+a^2b-a^3}\right)\cdot\left(1-\frac{b-1}{a}-\frac{b}{a^2}\right)=$$ $$= \left(\frac{a(a-b)}{b(a^2+b^2)}-\frac{2a^2}{(b-a)(b^2+ab+a^2)-ab(b-a)}\right)\cdot\frac{a^2-ab+a-b}{a^2}=$$ $$=\left(\frac{a-b}{b(a^2+b^2)}+\frac{2a}{(a-b)(a^2+b^2)}\right)\cdot\frac{(a-b)(a+1)}{a}=$$ $$=\frac{(a-b)^2+2ab}{b(a^2+b^2)}\cdot\frac{a+1}{a}=\frac{a+1}{ab}.$$