If x=a+b+\frac{(a-b)^{2}}{4a+4b} and y=\frac{a+b}{4}+\frac{ab}{a+b}, then what is the value of (x-a)^{2}-(y-b)^{2}?

Explanation

Simplifying the expressions: x-a = b + \frac{(a-b)^2}{4(a+b)} = \frac{4b(a+b) + (a-b)^2}{4(a+b)} = \frac{a^2+2ab+5b^2}{4(a+b)}. Similarly, y-b = \frac{a^2+2ab-3b^2}{4(a+b)}. Using the difference of squares, (x-a)^2 - (y-b)^2 = ((x-a)-(y-b))((x-a)+(y-b)). The difference is \frac{8b^2}{4(a+b)} = \frac{2b^2}{a+b}, and the sum is \frac{2(a+b)^2}{4(a+b)} = \frac{a+b}{2}. Their product is b^2.

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