‘.’ x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})
\frac{x}{y}= \frac{(a+b-\frac{ab}{a+b})}{(a-b+\frac{ab}{a-b})}
By, Componendo & Dividendo –
\frac{x+y}{x-y} = \frac{(a+b-\frac{ab}{a+b})+(a-b+\frac{ab}{a-b})}{(a+b-\frac{ab}{a+b})-(a-b+\frac{ab}{a-b})}
\frac{x+y}{x-y} = \frac{(2a + \frac{ab}{a-b} – \frac{ab}{a+b})}{(2b – \frac{ab}{a+b}- \frac{ab}{a-b})}
\frac{ x+y}{x-y} = \frac{(2a + \frac{a^2b+ab^2 – (a^2b-ab^2)}{a^2-b^2})}{(2b – \frac{(a^2b-ab^2 + a^2b+ab^2)}{ a^2-b^2})}
\frac{x+y}{x-y} = \frac{(2a + \frac{2ab^2}{a^2-b^2})}{2b – \frac{2a^2b}{a^2-b^2})}
\frac{ x+y}{x-y} = \frac{(\frac{2a^3 – 2ab^2 +2ab^2}{a^2-b^2})}{(\frac{2a^2b – 2b^3 – 2a^2b}{a^2-b^2})}
\frac{x+y}{x-y} = \frac{2a^3}{-2b^3}
So, x-y= -2b^3
Answer (A)