Reciprocators

Solving for $x$ : $\begin{aligned} \frac{a}{b}x = & \frac{b}{a} \\ x = & \frac{b^{2}}{a^{2}} \end{aligned}$

Solving for $y$ : $\begin{aligned} \frac{a}{b} + y = & \frac{b}{a} \\ \frac{a}{b} + y - \frac{a}{b} = & \frac{b}{a} - \frac{a}{b} \\ y = & \frac{b}{a} - \frac{a}{b} \\ y = & \frac{b^{2}-a^{2}}{ab} \end{aligned}$

Solving for $x+y$ : $\frac{b^{2}}{a^{2}} + \frac{b^{2}-a^{2}}{ab} = \frac{b^{3}}{a^{2}b} + \frac{ab^{2}-a^{3}}{a^{2}b} = \frac{-a^{3}+ab^{2}+b^{3}}{a^{2}b}$

$\implies \frac{-a^{3}+ab^{2}+b^{3}}{a^{2}b}$ is equal to $\boxed{-\frac{a^{3}-ab^{2}-b^{3}}{a^{2}b}}$ .