Does anyone know if a formula exists to obtain all solutions of the above Diophantine equation? All numbers integers.
Addendum: After seeing the answer from Tito Piezas III, I reconsidered the above equation and came up with an original solution that applies to two consecutive cubes when one of the cubes is equal to two squares. Solutions for three consecutive cubes exist, when the middle cube is equal to two squares, but I regret to say that I could not find a general solution for the third cube. $[(a^2+b^2) +1]^3= [a(a^2+b^2)+a]^2+[(a^2)b+b+b^3]^2+[a^2+b^2+1]^2$, all numbers positive integers and $a>b$.