For example $n=\sum_{k=1}^{100}10^{2^k}$.

I'm still struggling to see how that works. Could you explain please? Tx Merlin

Consider $P(x) = \sum_{k=1}^n x^{2^k}$. Clearly $P(1) = n$, and $P(1)^3 = n^3$. But $P(10)$ is the number proposed by Rust; its digits in decimal representations are $0$ and $1$, so the sum of its digits is $n$. When cubing $P(10)$, the coefficients of the powers of $10$ will not exceed $9$, so there is no carry-over, hence the sum of its digits is $n^3$. A simpler example would be $P(x) = \sum_{k=1}^n x^{3^k}$, where the digits of $P(10)^3$ can only be $0,1,3$ or $6$.