Insane Diophantine equation

It can be proved easily that $(x+3)^{3} > y^{3} > x^{3}$ Since both x and y are integers we have either $y = x+1$ or $y = x+2$ . Putting this into the equation we obtain $x = 0$ and $x = 9$ as solutions(Both occur when $y = x+2$ ).

Putting $x =0$ we get $y = 2$ .

Putting $x = 9$ we get $y = 11$ .

So the sum of all possible y-values = $2+11$ = $\boxed{13}$

Another solution:

y^3 = x^3 + 8x^2 - 6x + 8 = (x^3 + 6x^2 + 12x + 8) + (2x^2 - 18x) = (x + 2)^3 + 2x(x - 9).

Since the right-hand side has degree 3 with (x + 2)^3, the trivial solution here is to make the remaining terms (2x^2 - 18x) zero which is when x = 0 and x = 9 giving y = 2 and y = 11 respectively. We prove that there are no more solutions aside from these by using the gaps between perfect cubes.

Since (x + 1)^3 - x^3 = 3x^2 + 3x + 1, we compare 3x^2 + 3x + 1 > 2x^2 - 18x gives the inequality x^2 + 21x + 1 > 0 which is true for all positive integers. which implies no more solutions aside from y = 2 and y = 11.

If we put y = x + k (where k is a positive or a negative integer), we obtain a quadratic

x^2(8 - 3k) - x(6 + 3k^2) + (8 - k^3) = 0 with a discriminant

D = -3k^4 + 32k^3 + 36k^2 + 96k - 220.

In order to make D positive, we choose 1 < k > 12 and for k = 2 we have D = 18^2, which yields two solutions for y: 2 and 11. My approach to this problem wasn't nearly as straightforward as in the other two solutions, because it involved lots of calculation, but with this type of problems the first instinct is almost always to check the quadratic discriminant.