Simplify the expression: (x^2 - y^2) / (x^-2 - y^-2)?

(x² - y²) / (x⁻² - y⁻²)
= (x² - y²) / [(1 / x²) - (1 / y²)]
= (x² - y²) / [(y² / x²y²) - (x² / x²y²)]
= (x² - y²) / [(y² - x²) / x²y²]
= (x² - y²) / [-(x² - y²) / x²y²]
= (x² - y²) × [-x²y² / (x² - y²)]
= -x²y²
= -(xy)²

...... The answer is : D. -(xy)²

Put n = x^2 - y^2 = (x+y)(x-y). Put d = x^-2 -y^-2 = [(1/x) + (1/y)][(1/x) - (1/y)] = - (x+y)(x-y)/(xy)^2. Then
N/D = - (xy)^2 and choice D. pertains.

I would not tell you the answer but I can tell you how to do that.
You may know the identities of Algebraic expressions.
In the first term here it is x^2-x^2 which can be written as x^2-(x+y)100+xy
The second can be done as same. Goodluck it's easy to do.

(x^2)((x^2) - (y^2))/(1 - (x^2)(y^(-2)))

(y^2)(x^2)*((x^2) - (y^2))/((y^2) - (x^2))

-((xy)^2)

(x^2 - y^2) / (x^-2 - y^-2)
(x^2 - y^2) / [(y^2 - x^2) / x^2y^2]
x^2y^2(x^2 - y^2) / -(x^2 - y^2)
= -x^2y^2
= -(xy)^2
Answer choice:
D. -(xy)^2

(x^2 - y^2) / (x^-2 - y^-2)
= (x^2 - y^2) / (1/x^2 - 1/y^2) ---> use the power rules, a^-n = 1/a^n. The power becomes positive when it's in the denominator.
now cross multiply to get a common denominator x^2y^2
= (x^2 - y^2) / [(y^2 - x^2)/x^2y^2]
= (x^2 - y^2) / [-(x^2 - y^2) /x^2y^2] ---> factored out a negative so you can cancel the factors.
= [(x^2y^2)*(x^2 - y^2)]/-(x^2 - y^2) --> cancel common factors.
= -(x^2y^2)
= -(xy)^2

answer is D

x^-2 - y^-2 = 1/x^2 - 1/y^2 = (y^2- x^2)/(xy)^2 = -(x^2-y^2)/(xy)^2

(x^2- y^2)/[-(x^2-y^2)/(xy)^2)] = 1/(-1/(xy)^2) = -(xy)^2

Answer: D.

(x^2 - y^2)/(x^-2 - y^-2)

= (x^2 - y^2)/((1/x^2) - (1/y^2))

= (x^2 - y^2)/((y^2 - x^2)/(x^2y^2))

= (x^2 - y^2) * ((x^2y^2)/(y^2 - x^2))

= (x^2 - y^2) * ((x^2y^2)/-1(x^2 - y^2))

= (x^2y^2)/-1

= -x^2y^2

= -(xy)^2. The answer is D).

(x^2 - y^2) / (x^-2 - y^-2)
= (x² - y²) / (1/x² - 1/y²)
= (x² - y²) / [(y² - x²)/x²y²]
= (x² - y²) / [-(x² - y²) /x²y²]
= - x²y²
= -(xy)²

D

D

yes, i agree with above answer.
Ans::: -x^2 y^2