Really stuck on this problem. everyone i asked for is stuck too.
simplify
(x^-2 - y^-2) / (x^-1 - y^-1)

So negative exponents mean reciprocals so x^-1=1/(x^1)

x^-2=1/(x^2)
x^-2=1/(y^2)
so ([1/(x^2)]-[1/(y^2]) is the diffarence of two perfect squares which is (x^-2-y^-2)=(1/x-1/y)(1/x+1/y)

so
that equals (1/x-1/y)(1/x+1/y)

(x^-1)=1/x
y^-1=1/y
the bottom is (1/x-1/y)
so the equation is
[(1/x-1/y)(1/x+1/y)]/(1/x-1/y)
we can cancell out the (1/x-1/y) on the top and the bottom and be left with
(1/x-1/y)/1 or 1/x-1/y

The answer is the third choice