TITLE: Why are cubics separable over fields that are not characteristic 2 or 3
QUESTION [6 upvotes]: Why is it that cubics are separable over fields that are not of characteristic $2$ or $3$?
This is the starting point for some a discussion of the Galois group of a cubic, but I seem to be stuck right off the bat.

REPLY [4 votes]: You can actually do this "by hand" too. Our goal is to show that a monic irreducible polynomial $f(x)$ of degree $2$ or $3$ over $F$ splits into distinct linear factors in an algebraic closure $E$, as long as the characteristic of the $F$ is not $2$ or $3$. Suppose the irreducible $f(x)$ does not split into linear factors in $E$; we will get a contradiction. 
First, if the polynomial is of the form $(x - a)^3 = x^3 - 3ax^2 + 3a^2x + a^3$, then the $x^2$ coefficient $-3a$ must be in $F$. So $a$ is in $F$ too, using that $F$ is not of characteristic $3$.
The remaining case is that the polynomial is of the form $(x - a)^2(x - b)$.
For this we use the fact that for any polynomial $p(x)$, ${p'(x) \over p(x)} = \sum {1 \over x - r}$, where the sum is taken over the roots of $p(x)$. Applying this to $f(x)$ gives that ${f'(x) \over f(x)} = {2 \over (x - a)} + {1 \over x - b}$.  Note that $f(x) = x^3 - (b + 2a)x^2 + ...$, so that $b + 2a \in F$. Hence ${b + 2a \over 3} \in F$ too. Substituting this value of $x$ into ${f'(x) \over f(x)}$ gives gives ${9 \over 2(b - a)}$, which must be in $F$. Thus $b - a\in F$; note we use the characteristic is not $2$ or $3$ here. So $b = {2 \over 3}(b - a) + {1 \over 3}(b + 2a)$ is in $F$ as well, again contradicting the irreducibility of $f(x)$.