To "determine" all the cubic roots of $2$, with 2 being an element in $Z_7$? Cubic root of an element in a field?
Regarding to this question, I have clarified myself about the definition of a cubic root in a field. So the next question that pops up in my mind is that to find all cubic roots of some element in a field other than $\mathbb{Q}$ or so. I chose to "determine" all the cubic roots of $2$, with $2$ being the element of $\mathbb{Z}_7$
It can be easily seen that $p(x) = x^3 - 2$ is irreducible in $\mathbb{Z}_7[x]$. I construct the field extension $Z_7[x]/(x^3 - 2)$ of $\mathbb{Z}_7$ and denote $[q(x)] = q(x) + \langle x^3 - 2 \rangle$ for all $q(x) \in \mathbb{Z}_7[x]$ with degree less than or equal $2$.
Due to the way I construct the extension, naturally the polynomial $x^3 - [2]$ admits $[x]$ as a root. In other words, I found a cubic root of $2$. 
From this point, I meet the following problems:
1) It is not necessary that $\mathbb{Z}_7[x]/(x^3 - 2)$ be the splitting field of $p(x) = x^3 - 2$. So I may have to continue to construct extensions of that field to find all the cubic roots of $2$.
2) With the thoughts in 1), I try to determine the breakdown of $p(y) = y^3 - [2]$ in  $(\mathbb{Z}_7[x]/(x^3 - 2))[y]$. That means I aim to determine $r(y) \in (\mathbb{Z}_7[x]/(x^3 - 2))[y]$ of degree $2$ that satisfies
$$y^3 - [2] = (y - [x]).r(y) $$
Brute force will not work, since there are $(7^3)^2 = 117649$ monic polynomials with degree $2$  in $(\mathbb{Z}_7[x]/(x^3 - 2))[y]$.
How do I determine $r(y)$? If I know $r(y)$ and see whether it is reducible in $(\mathbb{Z}_7[x]/(x^3 - 2))[y]$ or not, in either cases I can point out the other two cubic roots of $2$.
Please give me some hints regard this problem. I am just an undergraduate who is studying the basics of field extension theories. I have minimal knowledge about splitting fields (how to prove the existence of one) and very little about Galois theory. Thank you for reading.
EDIT: I changed my way of using words and replace some notations to be more appropriate
 A: First a note on terminology; the element $[x]\in\Bbb{F}_7[x]/(x^3-2)$ is not a cube root of $2$ $\color{red}{\text{in }\Bbb{F}_7}$, but rather over $\Bbb{F}_7$. After all, it is not an element of $\Bbb{F}_7$. I also prefer the notation $\Bbb{F}_7$ in stead of $\Bbb{Z}_7$, as the latter is also used for $p$-adic integers.
To avoid confusion I'll write $\alpha:=[x]\in\Bbb{F}_7[x]/(x^3-2)$ so that $\Bbb{F}_7[\alpha]=\Bbb{F}_7[x]/(x^3-2)$  and $\alpha^3=2\in\Bbb{F}_7$.
To find other cube roots of $2$ you want to factor the polynomial $y^3-2\in(\Bbb{F}_7[\alpha])[y]$. By construction $\alpha$ is a root, so
$$y^3-2=(y-\alpha)(y^2+ay+b),$$
for some $a,b\in\Bbb{F}_7[\alpha]$. Expanding the right hand side and comparing coefficients shows that
$$a-\alpha=0,\qquad b-a\alpha=0,\qquad -\alpha b=-2,$$
from which it is immediate that $a=\alpha$ and $b=\alpha^2$. So now you want to factor the polynomial
$$y^2+ay+b=y^2+\alpha y+\alpha^2.$$
This is a quadratic polynomial, and as for $\Bbb{Q}$ the quadratic formula gives you the solutions:
$$y=\frac{-\alpha\pm\sqrt{\alpha^2-4\cdot1\cdot\alpha^2}}{2}=\frac{\alpha}{2}(-1\pm\sqrt{-3}).$$
Of course in $\Bbb{F}_7$ we have $\sqrt{-3}=\sqrt{4}=2$, and so the other cube roots of $2$ in $\Bbb{F}_7[\alpha]$ are $\frac{\alpha}{2}=4\alpha$ and $2\alpha$.

The above is a very mechanical (but also very certain) way of determining the other cube roots of $2$ in $\Bbb{F}_7[\alpha]$. Another way to find them is to note that $\Bbb{F}_7$ contains three cube roots of unity; we have $$1^3=2^3=4^3=1.$$
This means $\alpha^3=2$ implies $(2\alpha)^3=2$ and $(4\alpha)^3=2$, so the other cube roots of $2$ are $2\alpha$ and $4\alpha$.
