This question has no justification other than a bit of fun.


We all know that the cubic is solvable by "radicals" ($\root2\of{}$ and $\root3\of{}$) in characteristics $\neq2,3$. The formula was discovered by the Italians in the 16th century [(see here)](http://en.wikipedia.org/wiki/Cubic_equation).


In characteristic $2$, there should be a similar formula involving $\wp\_2^{-1}(\ )$ and $\root3\of{}$, and in characteristic $3$ there should be a formula involving $\root2\of{}$ and $\wp\_3^{-1}(\ )$. 


By $\wp\_2^{-1}(a)$ and $\wp\_3^{-1}(a)$ I mean a root of the polynomials $\wp\_2(T)=T^2-T-a$ and $\wp\_3(T)=T^3-T-a$ respectively, which give all cyclic extensions of degree $2$ and $3$ respectively. 


Has somebody worked out these formulæ ? 


**Edit**. I have accepted one of the answers --- the choice was difficult --- but I'm still curious as to whether these formulæ can be found somewhere in the literature.



Solving the cubic by "radicals" in characteristics 2 and 3
I asked an undergraduate (Dubravka Bodiroga at Hood College) to work these results out last summer. Here is her cubic formula in characteristic 3 (paraphrasing from something she sent me):


Consder the polynomial
$$
x^3 - a\_1 x^2 + a\_2 x - a\_3,
$$
where the coefficients belong to a commutative ring in which $3=0$. Assume moreover that $a\_1$ is invertible. Let $b$ be a solution to
$$
b^2 = -\frac{a\_2^3}{a\_1^6}+\frac{a\_2^2}{a\_1^4}-\frac{a\_3}{a\_1^3},
$$
and let $\beta$ be a solution to $y^3 - y - b = 0$. Then
$$
x^3 - a\_1 x^2 + a\_2 x - a\_3 = (x - (u\beta^2+v))(x-(u(\beta+1)^2+v))(x-(u(\beta+2)^2+v)),
$$
where $u=-a\_1$ and $v=a\_1 -\frac{a\_2}{a\_1}$.


If one inverts the procedure, letting
$$
x\_i = u(\beta+i)^2 + v,
$$
then ([Parson: assuming I haven't scrambled her indices])
$$
b= \frac{(x\_0 - x\_1)(x\_1 - x\_2)(x\_0 - x\_2)}{(x\_0+x\_1+x\_2)^3},
$$
and
$$
y= \frac{x\_2 + 2x\_1}{x\_0+x\_1+x\_2} = -\frac{2\times x\_2+1\times x\_1+0\times x\_0}{x\_0+x\_1+x\_2}.
$$


I believe she also worked out the (simpler) details for the cubic formula in characteristic $2$, but I could not find them just now. She used a heuristic method of Euler and B\'ezout to find the formulas, an exposition of which one can find in Tignol's book on Galois theory. She then solved for the auxiliary quantities $b$ and $y$ in terms of the $x\_i$ to see what Lagrange would have made of her solution procedure.


