When using a^2-b^2 = (a+b)(a-b)

In this case, you have

(x+y)^2 - (x-y)^2 = (x+y+x-y)*(x+y-x+y) = 4xy.

You get 2^4xy = 2^4 = 16

razorback wrote:Here I recognize that a possible concept is a^2 - b^2 = (a+b)(a-b). When I plugged the exponents into that format I got 2^1, for answer A, but was wrong! GMATPrep offers no explanations to its problems.


If xy = 1, what is the value of 2^(x+y)^2 all over 2 ^(x-y)^2

A) 2
B) 4
C) 8
D) 16
E) 32

OA: D

Problem says, 2^(x+y)^2 all over 2 ^(x-y)^2

Therefore, 2^(x^2 + y^2 + 2xy) all over 2^(x^2 + y^2 - 2xy)
2^(x^2 + y^2 + 2 * 1) all over 2^(x^2 + y^2 - 2 * 1) (because xy = 1)
2^(x^2 + y^2 + 2) all over 2^(x^2 + y^2 - 2)
2^(x^2 + y^2) * 2^2 all over 2^(x^2 + y^2) * 2^(-2) (Expanding)
2^2 all over 2^(-2) (2^(x^2 + y^2) got cancelled from numerator and denominator)
2^2 * 2^(2)
4 * 4
16

Hence, answer is D

edirik wrote:Or you can just say that x=y=1 and calculate the equation in 20 seconds and find that it is 16. Snce the answer choices are all numbers, it is easy to provide those numbers such as 1 to calculate the equation.

No. If x = 1/2, y = 2, then xy = 1 and x!=y.

edirik wrote:You are not right, I hope you really counted x = 1/2, y = 2 before you write your comment.

You can also use x = 1/2, y = 2 and see that the result is 16, as it is for x=1 and y=1.

But since we have x=1 and y=1, you do not need to select those numbers to avoid complex crunching.

Nevermind, I misunderstood what you originally said.