Basic field $\Bbb Z_7$ equation

Question

trying to figure if I understand the basics of $\Bbb Z_7$ field ${}=\{1,2,3,4,5,6\}$ on the following equation:

$6x^3$ = $5$

I tried to reduce the coefficients in the equation -

$6x^3$ = $5$ => $6*6x^3 = 6*5$ => $36x^3 = 30$ => mod7 => $x^3 = 2$

From this stage $x^3 = 2$. I don't know how to continue.

Thanks in advance.

Answer 1

Hint:

Clearly $0$ is not a solution. On the other hand, for any $x\ne0$, lil' Fermat says that $x^6=1$, so $(x^3)^2=1$. What can you deduce for $x^3$, knowing we're in a field?