How do I go about these proofs? I can't find much online about cubic reciprocity.
Suppose $p \geq 5$. If $p \equiv 1, 7 \pmod{12} $, then the number of distinct nonzero cubic residues (mod p) is $(p-1)/3$.
Suppose $p \geq 5$. $p$ does not divide $a$. $x^{3} \equiv a \pmod{p}$ has a solution r. Then $(x-r)(x^{2} + xr + r^{2}) \equiv 0 \pmod{p}.$ Show that $(x^{2} + xr + r^{2}) \equiv 0 \pmod{p}$ has 2 solutions not equal to r if and only if $p \equiv 1, 7 \pmod{12} $.