2011/068) Find all ordered pairs of integers (x,y) which satisfy x^3 + y^3 - 3x^2 + 6y^2 + 3x + 12y + 6 = 0

Rewrite this using x terms and y terms in groups

(x^3 - 3x^2 + 3x) + (y^3 + 6y^2 + 12y) = -6
==> (x^3 - 3x^2 + 3x - 1) + (y^3 + 6y^2 + 12y + 8) = -6 - 1 + 8
==> (x - 1)^3 + (y + 2)^3 = 1
as difference of 2 cubes cannot be 1 so one of them is zero and another is 1 and there is no other choice

x - 1 = 1 and y + 2 = 0 ==> (x, y) = (2, -2), or
x - 1 = 0 and y + 2 = 1 ==> (x, y) = (1, -1).

so solution sets are (2,-2) and (1,-1)