# 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?

• An alternative would be to calculate the symmetric functions for the new roots directly. No square roots involved, but still some calculations. – dxiv Nov 11 '16 at 23:56
• If you are more comfortable with identities, you can try the alternative suggested by @dxiv. Your method, although involving square roots does not have many intermediate side-steps. – Shraddheya Shendre Nov 12 '16 at 0:10
• Is the equation you obtained after substitution correct? I think it should start with $q^2/t^2$ and have an additional q on the right hand side. I might also be missing something. – Deniz Sargun Nov 12 '16 at 0:24
• Thank you for spotting that out -- yes I was a bit careless with my algebra! – Hugh Entwistle Nov 12 '16 at 0:37

Here is a hint for a way of solving it using elementary steps: Consider a degree 3 polynomial that have roots $\alpha, \beta, \gamma$:

$$(x-\alpha)(x-\beta)(x-\gamma),$$ this must then be the same polynomial as $$x^3+px+q.$$

Can you now relate now $\alpha, \beta, \gamma$ with $p$ and $q$? Do we get any relation between $\alpha, \beta, \gamma$?

Yes, from $$(x-\alpha)(x-\beta)(x-\gamma)=x^3+px+q$$ one obtains $\alpha \beta + \alpha \gamma + \beta\gamma= p$, $\alpha \beta \gamma =-q$, and $\alpha + \beta + \gamma=0$. Due to this latter relation, one can also write $(\alpha + \beta + \gamma)^2=0$, i.e., $\alpha^2 +\beta^2 +\gamma^2=-2(\alpha \beta + \alpha \gamma + \beta\gamma)=-2p$ and also $(\alpha \beta)^2 + (\alpha \gamma)^2 + (\beta \gamma)^2=(\alpha \beta + \alpha \gamma + \beta \gamma)^2-2(\alpha \beta\gamma)(\alpha +\beta+\gamma)=(\alpha \beta + \alpha \gamma + \beta \gamma)^2=p^2$

After this, consider a degree 3 function that has roots $\frac{\alpha \beta}{\gamma}, \frac{\alpha \gamma}{\beta}, \frac{\alpha \gamma}{\beta}$:

$$\left(x-\frac{\alpha \beta}{\gamma}\right)\left(x-\frac{\alpha \gamma}{\beta}\right)\left(x-\frac{\alpha \gamma}{\beta}\right)=\\ \frac{1}{\alpha \beta \gamma} \left[(\alpha \beta \gamma) x^3 - ((\alpha \beta)^2 + (\alpha \gamma)^2+(\beta \gamma)^2)x^2 + (\alpha \beta \gamma)(\alpha^2 + \beta^2 +\gamma^2)x -(\alpha\beta\gamma)^2\right]$$

Now use the obtained identities between $\alpha, \beta, \gamma$ and $p, q$ to write up the above degree 3 polynomial with the desired roots:

$$qx^3+p^2x^2-2pqx +q^2.$$

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