$x^3-y^2+ab=a-b(x+y)$

Question

Is it possible to find the set of solutions for this equation? I tried factorising but in vain: By adding and subtracting $x^2$ I get $$\left(x+\frac{b}{2}\right)^2-\left(y-\frac{b}{2}\right)^2=-x^3+x^2+a-ab$$ $$(x+y)(x-y+b)=-x^3+x^2+a-ab$$

Answer 1

Some solutions $(a,b,x,y)$:

(-3, -3, 1, 2)
(-3, -7, 2, 2)
(-9, -13, 1, 6)
(-3, -13, 3, 2)
(-9, -19, 2, 6)
(-3, -21, 4, 2)
(-9, -27, 3, 6)
(-3, -31, 5, 2)
(-9, -37, 4, 6)
(-3, -43, 6, 2)
(-9, -49, 5, 6)
(-3, -57, 7, 2)
(-9, -63, 6, 6)
(-3, -73, 8, 2)
(-9, -79, 7, 6)
(-3, -91, 9, 2)
(-9, -97, 8, 6)
(-3, -111, 10, 2)
(-9, -117, 9, 6)
(-3, -133, 11, 2)
(-9, -139, 10, 6)
(-3, -157, 12, 2)
(-9, -163, 11, 6)
(-3, -183, 13, 2)
(-9, -189, 12, 6)

(5, 7, 3, 13)
(19, 79, 3, 97)
(29, 5, 5, 19)
(29, 13, 13, 59)
(41, 7, 7, 29)
(41, 13, 3, 31)
(41, 17, 7, 43)
(47, 3, 3, 13)
(71, 11, 13, 61)
(73, 3, 17, 73)
(97, 3, 11, 41)
(97, 13, 7, 47)