Упростите выражение , help me please

A)
$\frac{(a-b)^{-2}+2(a^2-b^2)^{-1}+(a+b)^{-2}}{(a+b)^2+2(a^2-b^2)+(a-b)^2}= \frac{((a-b)^{-1}+(a+b)^{-1})^2}{((a+b)+(a-b))^2} = \\ \\ = \frac{( \frac{1}{a-b}+ \frac{1}{a+b} )^2}{(2a)^2}= \frac{( \frac{a+b+a-b}{(a-b)(a+b)} )^2}{(2a)^2} = \frac{(2a)^2}{(a^2-b^2)^2*(2a)^2} = \\ \\ = \frac{1}{(a^2-b^2)^2}=(a^2-b^2)^{-2}$

б)
$\frac{(x+y)(x-y)^{-1}+(x-y)(x+y)^{-1}-2}{(x+y)(x-y)^{-1}-(x-y)(x+y)^{-1}}= \frac{ \frac{x+y}{x-y}+ \frac{x-y}{x+y}-2 }{ \frac{x+y}{x-y}- \frac{x-y}{x+y} }= \\ \\ = \frac{ \frac{(x+y)^2+(x-y)^2-2(x-y)(x+y)}{(x-y)(x+y)} }{ \frac{(x+y)^2-(x-y)^2}{(x-y)(x+y)} }= \frac{(x+y-(x-y))^2}{(x+y-(x-y))(x+y+x-y)}= \\ \\ = \frac{(2y)^2}{2y*2x}= \frac{2y}{2x}= \frac{y}{x}$