The implied other side to the 1729 example, x^3 + y^3 = z^3 - 1 could also be thought of as a special case w^3 + x^3 + y^3 = z^3 (well, they both can, if the variables are integers). Since the more specific version has infinite examples, so must this. But what about if we bump those exponents up to 4? Is there a generalized Fermat's Last?
In fact every integer k is the sum of nine cubes, and every multiple of 6 is the sum of four cubes. Your equation is just the equation x^3 + y^3 + (-z)^3 + w^3 = 6k where k = 0.
The more general problem is known as Waring's problem. For example, every integer is the sum of 19 fourth powers. Of course special families of integers require fewer fourth powers.
Could this be further generalized to n integers raised to nth power added together can have a number raised to nth power integer, hence why there is no relationship such that x^3 + y^3 = z^3, but there are an infinite number of x^3 + y^3 + z^3 = a^3 (where z in this case can work as +-1).
