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).

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.