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?
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