Cubic non-residue calculation

Question

I am currently studying Cubic residue characters from Kenneth Ireland and Michael Rosen's "A Classical Introduction to Modern Number Theory", and this is the definition given in the book:

If $N(\pi) \neq 3$, the cubic residue character of $\alpha$ modulo $\pi$ is given by $(a)$ $ (\frac{\alpha}{\pi})_3 = 0$ if $\pi|\alpha$, and $(b)$ $ \alpha^{\frac{(N(\pi)-1)}{3}} \equiv (\frac{\alpha}{\pi})_3 \pmod \pi$ with $(\frac{\alpha}{\pi})_3$ equal to $1,\omega$, or $\omega^2$.

I am trying to solve an example. I took the Eisenstein prime to be 7. And now, I need to check the solvability of $x^3 \equiv \alpha \pmod 5$.

I have understood how to solve it in the case where $\alpha$ is a cube modulo 7. For example, $2^3 \equiv 3 \pmod 5$, which implies that 3 is a cube modulo 5. I want an $\alpha$ which is not a cube modulo 5 and then I also want to understand how we associate $\omega$ or $\omega^2$ values to that specific cubic residue character. Since I am working in the $Z[\omega]$ ring, this is confusing me.

$\textbf{Update 1:}$:

Using the exercises in chapter 9 of Kenneth Ireland and Michael Rosen's book, I have figured out how to check if $\omega$ and $(1-\omega)$ are cubic residues of a given primary prime:

(In this case, I took the primary prime to be 5.)

1. $(\frac{\omega}{5})_3=\omega^{2+0}$ where $5=(3m-1)+3n\omega \implies (3m-1)+3n\omega=5+0(\omega) \implies m=2, n=0.$ This calculation is based on the result of an exercise question which states that for a given primary prime $\gamma$, let $\gamma=a+b\omega$, and set $a=3m-1, b=3n$. then $(\frac{\omega}{\gamma})_3=\omega^{m+n}$.

2. From the $\textit{ Supplement to the Cubic Reciprocity Law}$, I figured out $(\frac{1-\omega}{5})_3=\omega^{2m}=\omega^{2 \cdot 2}=\omega^4=\omega$

$\textbf{Update 2:}$

If I had to take an element of the form $a+b\omega$ and check if its a cubic residue modulo 5, then would the following method be correct?

$(\frac{2-\omega}{5})_{3}=(2-\omega)^{\frac{25-1}{3}}=(2-\omega)^8$ (by the definition of cubic residue character)

Do i just calculate $(2-\omega)^8$ and that is my result? (Upon calculation, I got the value to be 0. This means that $5|2-\omega$. To verify this I need to find an element $a+b\omega$ such that $5(a+b\omega)=(2-\omega)$. I am not able to find that.)