Quote by Norwegian
You can probably generalize further, and replace the 1 in 2x^{3}1 with any cube.

Yeah! For any cube(r^3) and any integer(n), the quadruple of real numbers of the form {n,n,0,r} satisfies the equation x^3+y^3+z^3+t^3=r^3.
EDIT:Sorry, I thought you meant replace the 2x^31 by any cube. In order to generalize to the case you mentioned above, in the quadruple of real numbers {10+n,10n,(60n^2)^1/3,1}, just replace the 1 by any integer, and you get infinitely many representations of the number 2x^3f^3 as the sum of four cubes.