33 = 88661289752875283 + (-8778405442862239)3 + (-2736111468807040)3
That result was discovered by Andrew Booker of the University of Bristol just this year.
It leaves 42 as the only positive integer less than 100 that has not been represented as the sum of three cubes.
(We can omit numbers that give a remainder of 4 or 5 when divided by 9, since those are known to be ineligible. But can every other integer be expressed in this way? It’s an open problem.)
(Thanks, Kate.)
09/06/2019 UPDATE: The case of 42 has now been solved, by Andrew Booker at Bristol and Andrew Sutherland at MIT:
42 = (-80538738812075974)3 + 804357581458175153 + 126021232973356313
The lowest unsolved case is now 114.