Question 549954
From the binomial expansion, we have


*[tex \LARGE (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3], which we can write as


*[tex \LARGE (x+y)^3 = x^3 + 3xy(x+y) + y^3].


Clearly, all (x,y) satisfying x+y = 1 work. Note that the other tutor also stated x+y = 1, using a valid argument, but the solution is much longer.


Now suppose x+y was not equal to 1. We may treat the original equation as a cubic equation by fixing a value for y and solving for x. In other words, we want to find integer roots for the cubic equation


*[tex \LARGE x^3 + 3xy + (y^3 - 1) = 0] other than x = 1-y.


Here, we use the fact that if r is a root of a polynomial, then 1-r is a factor. We divide both sides by x - (1-y) or x+y-1:


*[tex \LARGE x^2 + x(1-y) + (y^2 + y + 1) = 0]. Solve for x via the quadratic formula.


*[tex \LARGE x = \frac{y-1 \pm \sqrt{(y^2 - 2y + 1) - 4(y^2 + y + 1)}}{2}]


*[tex \LARGE = \frac{y-1 \pm \sqrt{-3y^2 - 6y - 3}}{2}]


*[tex \LARGE = \frac{y-1 \pm \sqrt{-3(y^2 + 2y + 1)}}{2}]


One thing to note: the expression inside the radical is equal to -3(y+1)^2. Since the square of a number is always non-negative, we will never have a real root, unless the expression inside the radical is equal to zero. Hence, we want y to equal -1. If y = -1, then x is also equal to -1. It is seen that (x,y) = (-1,-1) satisfies the original equation.


Therefore, the only solutions to the original equation are ordered pairs (x,y) satisfying x+y = 1 or (-1,-1). We want to evaluate x-y, which is equal to (x+y)-2y, or 1-2y, which can be any odd number (since y can equal anything). Also, (-1,-1) yields x-y = 0. Hence, the possible values for x-y are odd numbers and 0.