"completing the cube" Given a quadratic polynomial $p(z) = z^2 +a z +b$ in $\mathbb{C}[x]$, we can "complete the square" to write $p(z) = g(f(z)^2)$ where $f,g$ are translations.
\begin{align*}
f(z) = z + a/2 && g(z) = z+ b - a^2/4
\end{align*}
In particular, $f,g$ are entire diffeomorphisms homeomorphisms of $\mathbb{C}$.

Question: Given a monic degree 3 polynomial $p(z)$ with complex coefficients, is it always possible to find entire diffeomorphisms homeomorphisms $f,g$ of $\mathbb{C}$ such that $p(z) = g( f(z)^3)$?

It the answer is yes, how explicit can the maps $f,g$ be? Does this work for higher degree polynomials?
 A: OK I thought about it some more and I think the answer is no. 

Easy fact: Let $q : \mathbb{C} \to \mathbb{C}$ be a function, let $f,g$ be bijections of $\mathbb{C}$ and let $c \in \mathbb{C}$. Then $g(q(f(z))) = g(c)$ has as many solutions as $q(z) = c$. In fact, $f$ puts the solution sets in bijection.

Now, notice that $z^3 = 0$ has precisely one solution. That means that, for any function of the form $p(z) = g(f(z)^3)$ where $f,g$ are bijections, there should exist a $c \in \mathbb{C}$ such that $p(z) = c$ has exactly one solution. 
However the above does not occur for all degree three polynomials. Consider for instance $p(z) = z^3 + 3z^2$. Let $c \in \mathbb{C}$. The equation $p(z) = c$ will have three solutions unless the discriminant $\Delta_c = 27(4-c)c$ of $p(z) - c$ vanishes. So $p(z) = c$ only has less than three solutions when $c=0$ or $4$. In each of these cases, $p(z) = c$ has two solutions. Thus, it is not even possible to find discontinuous bijections $f,g : \mathbb{C} \to \mathbb{C}$ such that $g(f(z)^3) = z^3 + 3z^2$. 
