There's Cubic Residues As Well?

Relevant wiki: Chinese Remainder Theorem

Since $2016=7\times 9\times 32$ , we can solve the congruence $x^3\equiv 1$ modulo 7, 9, and 32 and then multiply the numbers of solutions, by the Chinese Remainder Theorem.

While there is a general theory, the numbers are so small that we can easily do this "by inspection".

Modulo 7 we find the solutions $x\equiv1,2,4$ , and modulo 9 we have $x\equiv1,4,7$ .

Note that $x^8\equiv 1 \pmod{32}$ for odd $x$ . Thus, if $x^3\equiv 1\pmod{32}$ , then $x^9\equiv x\equiv 1 \pmod{32}$ , so that $x\equiv1$ is the only solution in this case.

The total number of solutions is $3\times 3\times 1=\boxed{9}$