Does the equation $x^3-9=0$ have any solutions in $(\mathbb{Z}/31 \mathbb{Z})$? This is my first time "solving an equation" in a group so I feel like that may be the source of my troubles. Here's what I have so far, although I'm not sure how to progress:

Assume $c$ is a solution to $x^3-9=0$ in $(\mathbb{Z}/31 \mathbb{Z})$.
  We then have $$c^3-9=0\implies c^3=9 \implies |c^3|=|9|$$ and since on
  the LHS we're multiplying elements in $(\mathbb{Z}/31 \mathbb{Z})$, we
  consider the order of $[9]_{31}$ in $(\mathbb{Z}/31 \mathbb{Z})^*$
  which is $15$ since $$9^{15} \equiv 1 \ \text{mod} \ 31.$$ Thus
  $|c^3|=15$.

I feel like I should be able to deduce $|c|$ from $|c^3|$ but it doesn't seem to be clear to me. Could someone assist me in figuring out how to get $|c|$? 
From there, my next steps would be to ensure it divides $31$. Since $(\mathbb{Z}/31 \mathbb{Z})^*$ is abelian and cyclic thus has an element of maximal finite order of $31$, and if $|c| \leq 31$ then $|c|$ must divide $31.$
 A: There is no solution as
$$c^{30} = 9^{10} = 5$$
but $$c^{30} = 1 \ \ \forall c \not =0$$
A: If $x^3=9$ then $x^{15}=9^5$. So $9^5\equiv 1\pmod{31}$. (Since $9^5\not\equiv -1$ , or $9^{15}\equiv -1$ and hence $9$ would not be a square modulo $31$.)
So you only need to check if $9^5\equiv 1\pmod{31}$.
A: Another approach uses a pigeonhole argument.
We have $\bmod 31$:
$2^3\equiv 8, (-3)^3\equiv 4=8/2$
Then $2$ is a cubic residue and so are all powers of $2$ and their negatives, thus:
$\pm 1, \pm 2, \pm 4, \pm 8, \pm 16$
That is $10$ nonzero residues, whereas the total number of cubic nonzero residues must be $(31-1)/3=10$.  Perforce there is no possibility left for $9$ or any other nonzero residue to be a cubic one.
As a corrollary, $3$ shall be a primitive root since this cannot be cubic (above), quadratic (QR), or fifth power ($3^3 \not \equiv \pm 1$).
A: All numbers $a$ coprime to $31$ satisfy $a^{30}\equiv 1 \bmod 31$. There is a primitive root $g$ such that $k=30$ is the smallest positive value such that $g^k\equiv 1 \bmod 31$, so any number coprime to $31$ can be expressed as equivalent to a power of $g$. This means that for $a^6$, in particular, there are only $5$ distinct values possible, since $a^6\in\equiv \{g^6,g^{12},g^{18},g^{24},g^{30}\equiv 1\}$
We can also see that $2^5\equiv 1 \bmod 31$, so also $\{4^5, 8^5, 16^5\}\equiv  1 \bmod 31$. $30/6 = 5$ and thus $e=\{1,2,4, 8, 16\} $ are the $5$ possible values that can have $a^6\equiv e \bmod 31$. 
Obviously $9$ is a square, so if $9$ is also a cube $(c^3\equiv 9 \bmod 31)$ then it must also be a sixth power $(d^6\equiv 9 \bmod 31)$ .
Thus  $(c^3\equiv 9 \bmod 31)$ has no solutions $c$.
A: Slightly more efficiently, we get that if $c^3\equiv 9\equiv 3^2\pmod{31}$, then $c^{15}\equiv 3^{10}\pmod{31}$.
$$3^{10}\equiv (27)^3\cdot 3$$
$$\equiv (-4)^3\cdot 3\equiv -64\cdot 3$$
$$\equiv -2\cdot 3\equiv -6\pmod{31},$$
Since $\left(c^{15}\right)^2\equiv 1\pmod{31}$ for all $c\in\mathbb Z$ such that $31\not\mid c$ by Fermat's little theorem,
we get $31\mid \left(c^{15}+1\right)\left(c^{15}-1\right)$,
so $31\mid c^{15}+1$ or $31\mid c^{15}-1$ by Euclid's lemma,
so $c^{15}\equiv \pm 1\pmod{31}$, contradiction.
((In fact, $a^{\frac{p-1}{2}}\equiv \left(\frac{a}{p}\right)\pmod{p}$ for all $a\in\mathbb Z$, odd primes $p$ by Euler's criterion))
((and also, in this case, $c^{15}\equiv 1\pmod{31}$, because
if $c^{15}\equiv -1\pmod{31}$,
then $\left(3^5\right)^2\equiv -1\pmod{31}$,
contradiction, because $-1$ isn't a quadratic residue mod $31=4\cdot 7+3$ by Quadratic Reciprocity))
