Showing $x^3-b$ is ireducible over $F_k$ given some conditions I am self learning some number theory, and came upon this question which i cannot solve. Can someone give me a solution to thie, along with an explanation? I have been going around in circles and coming up with nothing usefull. 
Let k be a prime power and let $b$ be an element of $(F_k)$*. Show that the polynomial $x^3 - b$ is ireducible over $F_k$ iff $3|(k-1)$ but 3 does not divide $(k-1)/order(b)$
 A: Suppose $3$ does not divide $k-1$. Then there exist integers $s$ and $t$ such that $3s+(k-1)t=1$. It follows that 
$$b=b^1=(b^s)^3(b^{k-1})^t= (b^s)^3.$$
So $x^3-b$ has a root in $F_k^\ast$, namely $b^s$, and is therefore not irreducible.
Suppose now that $3$ divides $k-1$.  Let $e$ be the order of $b$, and  suppose that $3$ divides $\frac{k-1}{e}$. Let $g$ be a generator of $F_k^\ast$, and let $b=g^m$. Then $g^{me}=1$, so $k-1$ divides $me$. Since $3$ divides $\frac{k-1}{e}$, it follows that $3$ divides $m$, say $m=3l$, and therefore  $b=(g^l)^3$, so $x^3-b$ is reducible.
Finally, we show that if $3$ divides $k-1$, but  $3$ does not divide $\frac{k-1}{e}$, then $x^3-b$ is irreducible. 
Suppose to the contrary that $x^3-b$ is reducible over the field. So it can be expressed as a product of two polynomials of degree less than $3$. One of the polynomials then has degree $1$. Without loss of generality we may assume it is of the form $x-a$. Then $a^3=b$, and therefore $a^{3e}=1$. So $3e$ divides $k-1$, contradicting the fact that $3$ does not divide $\frac{k-1}{e}$.
