(In reply to

Analytic attempt...? by tomarken)

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

It also eliminates the need for some of the "brute forcish" methods he used to eliminate 89, 94, 97, 98, 68, and 92.