Hint: Prove that if $n=k^3$ for some integer $k$, then the number of divisors of $n$ is one more than a multiple of $3$. (E.g., $2^3$ has $4$ divisors, $6^3$ has $16$ divisors, $100^3$ has $49$ divisors, ...) Since $999$ is not one more than a multiple of $3$, if $n$ has $999$ divisors then it cannot equal $k^3$ for any integer $k$.
To prove this, consider the prime decomposition of $k$. We can write
for some distinct primes $p_1,\dots,p_t$. Then a factor of $k^3$ looks like
where each $\beta_i$ satisfies $0 \leq \beta_i\leq 3\alpha_t$. How many such factors are possible?