<s>[INST] Simplification of rational expression gone wrong(high school math) Currently doing high school math and can't get this one right. I think I'm using entirely incorrect practices and am trying to pinpoint what it is. Could someone tell me where exactly I went wrong?
Original expression:
$$\left(\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}\right)\times \left(\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}\right)^{-1}$$
When I simplify the equations in parentheses I get:
$$\frac{b^2-ab+a^2}{ab(a+b)}\times\frac{1}{a^{2}}$$
Which should result in:
$$\frac{b^2+a(a-b)}{a^3b(a+b)}=\frac{b^2-a}{a^3b}$$
That is of course not right. The correct answer according to the book is $\frac{b-a}{a}$
 [/INST] Go slowly. Do it in blocks.
First block:
\begin{align}
\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}
&=\frac{b}{a(a+b)}-\frac{b-a}{b(a+b)} \\[4px]
&=\frac{b^2-a(b-a)}{ab(a+b)} \\[4px]
&=\frac{a^2-ab+b^2}{ab(a+b)}
\end{align}
Second block:
\begin{align}
\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}
&=\frac{a^2}{b(b^2-a^2)}+\frac{1}{a+b} \\[4px]
&=\frac{a^2}{b(b-a)(b+a)}+\frac{1}{a+b} \\[4px]
&=\frac{a^2+b(b-a)}{b(b-a)(b+a)} \\[4px]
&=\frac{a^2-ab+b^2}{b(b-a)(b+a)}
\end{align}
Putting the pieces together:
$$
\left(\frac{b}{a^{2}+ab}-\frac{b-a}{b^2+ab}\right)\times \left(\frac{a^2}{b^3-a^2b}+\frac{1}{a+b}\right)^{-1}=
\frac{a^2-ab+b^2}{ab(a+b)}\frac{b(b-a)(b+a)}{a^2-ab+b^2}=
\frac{b-a}{a}
$$
Now you see that you managed the first block correctly, but failed in the second one. Check carefully your steps and you'll see where you went astray.
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