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 $x=u-\frac mu=v-\frac mv$).

Your property is indeed true except possibly when $\gamma=0$, i.e. when $x^3+px+q$ has a double root. 
If $\gamma\neq 0$, taking $\Delta=\frac{(mx^2-nx+2m^2)}{\gamma}$
you are done; notice the identity
$$ (mx^2-nx+2m^2)^2=(n^2+m^3)(x^2+4m)+m(mx-2n)(x^3+px+q).$$
If $\gamma=0$ then $n=-\tau^3\neq 0$ by hypothesis, and $x^3+px+q$ can be rewritten as
$x^3-3\tau^2x-2\tau^3=(x+\tau)^2(x-2\tau)$. If $x=2\tau$, your property is true with $\omega=1$. If $x=-\tau$ however, your property is true iff $F$ contains a primitive third root of unity. This is false when $F={\mathbb Q}$ and $\tau=1$ for example.
