Transformation of Roots for a Cubic Polynomial

Consider $$x^3 + px + q = 0$$
with roots $\alpha, \beta. \gamma$

We want to find the degree 3 polynomial that has roots :
$$\frac{\alpha\beta}{\gamma}, \frac{ \alpha\gamma}{\beta}, \frac{ \beta\gamma }{\alpha}$$

My attempt so far:

$$\frac{ \alpha\beta}{\gamma} = \frac{\alpha\beta\gamma}{\gamma^2} = - \frac{q}{\gamma^2}$$
and similarly for all of the others:

So now I let $$t = - \frac{q}{x^2}$$ and substitute into the original polynomial; except I will multiply everything by $x$ first to make my substitution easier.

$$x^4 + px^2 = -qx$$
$$\frac{q^2}{t^2} - \frac{pq}{t} = - q\sqrt{ -\frac{q}{t} }$$
Squaring both sides and multiplying by $t^4$, we obtain:
$$qt^3 +p^2t^2 -2pqt +q^2 = 0$$

Except this seems like a messy, tedious solution as you have to muck around with square roots etc. Is there a nicer way to do this?