Solution
Clearly 0 satisfies the problem statement, and negative numbers cannot satisfy. So now we may assume that the integer is positive.
Suppose $N = a_1 \dots a_d$ where each $a_i$ is a single digit, and $a_1 \neq 0$. Because $N \geq {10}^{d-1}$ and $N = (a_1 + \dots + a_d)^3 \leq (9d)^3$ so we must have: $$10^{d-1} \leq 9d^3 $$
For large $d$ this inequality fails to hold. In fact, it's only true for $d \leq 4$. So $N$ is a 4-digit number, which is also a cube. Therefore $N \leq 21^3$ because $22^3 > 10^5$.
Now consider modulo 9: we know that $N \equiv (a_1 + \dots + a_d) \mod 9$ so: $$ N = (a_1 + \dots + a_d)^3 \equiv N^3 \mod 9$$ Among all the residues modulo 9, only 0, 1, 8 satisfy $x \equiv x^3 \mod 9$. Therefore $N \equiv 0, 1, 8 \mod 9$.
So now our candidates are: $N = 1^3, 8^3, 10^3, 17^3, 18^3$. By inspection, only $N = 1^3 = 1, 8^3 = 512, 17^3 = 4913, 18^3 = 5832$ satisfy.
