[Math] Transformation of Roots for a Cubic Polynomial

Question

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?

Answer 1

You just have to express the elementary symmetric functions of $\;\dfrac{\alpha\beta}\gamma, \dfrac{\beta\gamma}\alpha$ and $\dfrac{\gamma\alpha}{\beta}$ with $p, q,r$, then use Vieta's relations.

• $\begin{aligned}[t]\dfrac{\alpha\beta}\gamma+ \dfrac{\beta\gamma}\alpha+\dfrac{\gamma\alpha}{\beta}&=\dfrac{(\alpha\beta)^2+(\beta\gamma)^2+(\gamma\alpha)^2}{\alpha\beta\gamma}=\dfrac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\beta\gamma^2)}{\alpha\beta\gamma}\\&=\dfrac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2}{\alpha\beta\gamma}-2(\alpha+\beta+\gamma)=\frac{p^2}{-r}\end{aligned} $
• $\dfrac{\alpha\beta}\gamma\cdot\dfrac{\beta\gamma}\alpha+\dfrac{\beta\gamma}\alpha\cdot\dfrac{\gamma\alpha}{\beta}+\dfrac{\gamma\alpha}{\beta}\cdot\dfrac{\alpha\beta}\gamma=\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=-2p$
• $\dfrac{\alpha\beta}\gamma\cdot\dfrac{\beta\gamma}\alpha\cdot\dfrac{\gamma\alpha}{\beta}=\alpha\beta\gamma=-r$

Hence the equation is $$x^3+\frac{p^2}r x^2-2px+r=0\iff rx^3+p^2x^2-2prx+r^2=0.$$