For all numbers x and y, the operation # is defined by x#y = (x + y)^3

We want a point at which

\((x+y)^3 = (x-y)^3\)

put in the values of x and y from the options.
The expression holds true when x=-2 and y=0

Answer:- D

For all numbers x and y, the operation # is defined by

x # y = (x + y)^3,

whereas the operation ~ is defined by

x ~ y = (x - y)^3.

At which of the following points in the xy plane is the value of x # y equal to the value of x ~ y?

(A) (1, 1)
(B) (1, -1)
(C) (0, 2)
(D) (-2, 0)
(E) (-2, 2)

let (x + y)^3 = LHS and (x - y)^3 = RHS

Substitute the values in LHS and RHS resp.
A . LHS = 8 not equals RHS =0

B. LHS = 0 not equals RHS =8

C. LHS = 8 not equals RHS =-8

D. LHS = -8 not equals RHS =-8

E. LHS = 0 not equals RHS

Hence D ans

MANHATTAN GMAT OFFICIAL SOLUTION:

To attack this problem, we need to deal with two “strange symbol” formulas. We need to identify which of the five points given in the answer choices makes the two formulas equal.

One lengthy approach is to set the formulas equal to each other and perform algebra.

x # y = x ~ y

\((x + y)^3 = (x-y)^3\)

\(x^3+ 3 x^2y + 3yx^2 + y^3 = x^3-3{x^2}y + 3y{x^2}-y^3\)

(You don’t have to know that expansion – you can get it by FOILing)

\(3{x^2}y +y^3 = -3{x^2}y -y^3\)

\(6{x^2} y + 2 y^3 = 0\)

\(y(6x^2+2y^2) = 0\)

Either y = 0 or \(6x^2 + 2y^2 = 0\). The latter is only true if both x and y equal 0. So we must have at least the condition that y = 0. Comparing to the choices, we should see that only (-2, 0) works.

Alternatively, by looking at the two formulas, we might search for a quick shortcut. When would these expressions be equal? By asking ourselves that question, we might realize that if y = 0, both formulas reduce to x^3. So the two formulas will be equal whenever y= 0. The only viable option among the five answer choices is (-2, 0).

Finally, we can just “plug and chug” the answer choices. This might take more time, but as long as our substitutions and computations are correct, we will get the right answer.

The correct answer is D.

Simply set the two equations equal to each other

(x-y)^3 = (x+y)^3 - we should also know that a negative number to the third power will always be negative so as to save time in our calculations

(-2 +0)^3 = (-2+0)^3

Thus "D"

Given that x # y = (x + y)^3 and x ~ y = (x - y)^3 and we need the point at which x # y equal to the value of x ~ y

x # y = x ~ y
=> (x + y)^3 = (x - y)^3

Taking cube root both the sides we get

x + y = x - y
=> 2y = 0
=> y = 0

Only option choice with y = 0 is (-2,0)

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters