How to prove the cubic formula without root extraction I'm trying to prove the cubic formula, in the following form:

Given a field $F$ and $x,p,q\in F$, define $m=\frac p3$ and $n=\frac q2$, and suppose also that $\gamma,\tau$ are given such that $\gamma^2=n^2+m^3$ and $\tau^3=\gamma-n\ne0$. Then $x^3+px+q=0$ if and only if $x=\omega\tau-\frac m{\omega\tau}$ for some $\omega$ satisfying $\omega^3=1$.

(I have already reduced the statement from the more general $ax^3+bx^2+cx+d=0$ by a linear transformation.) The usage of $\gamma$ and $\tau$ substitutes for the extraction of square and cube roots, and allows me to stick to just field axioms in the proof - ideally I want this proof to work for any field with characteristic other than $2$ or $3$.
But making this approach work with Cardano's method is giving me some issues. Since the reverse implication is just a matter of algebra with the given expressions of $\tau,\gamma$, suppose that we are given $x$ such that $x^3+px+q=0$; we want to find some root of unity $\omega$ satisfying $x=\omega\tau-\frac m{\omega\tau}$. The key of Cardano's method is to observe that the conditions $u+v=x$, $uv+m=0$ together with the condition on $x$ yields a quadratic polynomial $z^2+qz-m^3=0$ whose roots are $u^3$, $v^3$, and solving this with the quadratic equation gives something amenable to a cube root extraction.
But why can we choose such $u,v$ in the first place? Equivalently, why should we expect a priori that the "substitution" $x=u-\frac mu$ has a solution in $u$? It follows from the quadratic formula, with discriminant $x^2+4m$, but I see no reason to believe that this is a square given only $\gamma$ and $\tau$.
If we assume that there is a $\Delta^2=x^2+4m$, then $u=\frac x2+\Delta$ and $v=\frac x2-\Delta$ solve the equations, so we can proceed as normal: $u^6+qu^3-m^3=0$, so by the quadratic formula with discriminant $q^2+4m^3=(2\gamma)^2$, we get $u^3=-n\pm\gamma$ and $v^3=-n\mp\gamma$; in one case we have $u^3=\gamma-n=\tau^3$ so $\frac u\tau$ is the desired root of unity, and in the other case $\frac v\tau$ is the desired root of unity (since $x=u-\frac mu=v-\frac mv$).

$x^3+px+q$ does have at most three roots in the algebraic closure of $F$.
Meanwhile, given the existence of $\gamma$ and $\tau$, and if necessary taking $\omega$ in the algebraic closure of $F$, we can just verify that each of $\tau-\frac{m}{\tau}$, $\omega\tau-\frac{m}{\omega\tau}$, and $\omega^2\tau-\frac{m}{\omega^2\tau}$ are roots. Characterizing all three at once as $\omega^i\tau-\frac{m}{\omega^i\tau}$, we can see:
$$
\begin{align}
&\left(\omega^i\tau-\frac{m}{\omega^i\tau}\right)^3+p\left(\omega^i\tau-\frac{m}{\omega^i\tau}\right)+q\\
&=\tau^3-3m\omega^i\tau+3\frac{m^2}{\omega^i\tau}-\frac{m^3}{\tau^3}+p\omega^i\tau-p\frac{m}{\omega^i\tau}+q\\
&=\tau^3-p\omega^i\tau+\frac{mp}{\omega^i\tau}-\frac{m^3}{\tau^3}+p\omega^i\tau-p\frac{m}{\omega^i\tau}+q\\
&=\tau^3-\frac{m^3}{\tau^3}+q\\
&=\gamma-n-\frac{m^3}{\gamma-n}+q\\
&=\gamma-n-\frac{m^3(\gamma+n)}{\gamma^2-n^2}+q\\
&=\gamma-n-\frac{m^3(\gamma+n)}{m^3}+q\\
&=\gamma-n-(\gamma+n)+q\\
&=-2n+q\\
&=0
\end{align}
$$
Assume for the moment that the three roots that were just demonstrated are distinct. Then, if  you had an $x$ that was a root of $x^3+px+q$, whether $x$ was in $F$ or not, it has to equal one of these three demonstrated roots, which are each of the form $$(\text{cubic root of $1$})\tau-\frac{m}{(\text{cubic root of $1$})\tau}$$ as required. 
If the three demonstrated roots are not distinct, there is a little more to think through.
