We start off by considering . Let , therefore since , we arrive at from basic simplification. Therefore, we now get or . We look at the two following cases: and .

Note that . From here, we have



Therefore, we have when and when , or when and when .

Summary of findings so far:
The equation has a solution when , in fact letting we have , which when there exists a , also works, therefore we have to test which gives us absurd.
The equation has no solutions when .

Therefore, when has two solutions, this implies that the equation is solved by three numbers in some order. Therefore, besides for , we have among the numbers, they are divided into groups of three's in terms of the cubic residues. By this, I mean that if , then so does and , so we count each residue in groups of threes. Hence, we have residues when , or therefore .

When has zero solutions when , this implies that . Therefore, all values are distinct, and we get .

When , we get so .


Therefore, as final counts for the number of cubic residues (including 0), we get
.