The sum of all digits from 0 through 9 is 0 mod 9, so n cannot be 1,2,4,5,7,8 mod 3, since n^2 + n^3 would not be 0 mod 9. n must therefore be divisible by 3. That gets tomarken's list down to
54
63
69
72
84
87
It also eliminates the need for some of the "brute forcish" methods he used to eliminate 89, 94, 97, 98, 68, and 92.
re: Analytic attempt...?
