Problem Solving

For any operation ? that acts on two numbers x and y, the commutator is defined as x?y – y?x. For which of the following operations is the commutator not equal to zero for some values of x and y?

I. x?y = xy
II. x?y = (x – y)^2
III. x?y = x^3 + 3x^2y + 3xy^2 + y^3

(A) I only
(B) II only
(C) III only
(D) II and III only
(E) I, II, and III

I don't find the answer among the choices available

I x?y = xy

so commutator = xy - yx so always 0

II x?y = (x-y)^2

so commutator = (x-y)^2 - (y-x)^2 taking -1 in one and squaring again makes it 1 so always 0

III x?y = (x+y)^3

so commutator = (x+y)^3 - (y-x)^3 its cube so -1 common wont help so not always 0

so IMO ans = C

OA Plz.

aspiregmat wrote:I x?y = xy

so commutator = xy - yx so always 0

II x?y = (x-y)^2

so commutator = (x-y)^2 - (y-x)^2 taking -1 in one and squaring again makes it 1 so always 0

III x?y = (x+y)^3

so commutator = (x+y)^3 - (y-x)^3 its cube so -1 common wont help so not always 0

so IMO ans = C

OA Plz.

can u pl. explain how do u get the highlighted term? I mean why it is (y-x)^3 and not (y+x)^3

real2008 wrote:

aspiregmat wrote:I x?y = xy

so commutator = xy - yx so always 0

II x?y = (x-y)^2

so commutator = (x-y)^2 - (y-x)^2 taking -1 in one and squaring again makes it 1 so always 0

III x?y = (x+y)^3

so commutator = (x+y)^3 - (y-x)^3 its cube so -1 common wont help so not always 0

so IMO ans = C

OA Plz.

can u pl. explain how do u get the highlighted term? I mean why it is (y-x)^3 and not (y+x)^3

My mistake ...

it should be (x+y)^3 - (y+x)^3 so Ans IMO none of the above , which is not in the ans choice....

I also thought the answers should be none. This is a Manhattan Gmat challenge question I found on MBA mission. Below is the exp if anyone can make sense of it please share.


https://www.mbamission.com/blog/2009/08/ ... llenge-63/