, 2016

Solution:
Since cubes are $0$ or $\pm 1$ modulo $9$, by inspection we see that we must have $a^{3} \equiv b^{3} \equiv 0 \pmod{3}$ for this to be possible. Thus $a$, $b$ are divisible by $3$. But then we get $3^{3} \mid 2016$, which is a contradiction.

One can also solve the problem in the same manner by taking modulo $7$, since all cubes are $0$ or $\pm 1$ modulo $7$. The proof can be copied literally, noting that $7 \mid 2016$ but $7^{3} \nmid 2016$.