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If $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4+px^3+qx^2+rx+s=0$, Show that the equation whose roots are $(\alpha \beta + \gamma \delta), (\beta \gamma + \alpha \delta), (\gamma \alpha + \beta \delta)$ is $$x^3-qx^2+(pr-4s)x-(r^2-4qs+p^2s)=0$$

I'm really stuck on this one. I've tried using simple manipulations, methods for finding equations with symmetric functions of roots, Newton's theorem. But, nothing seems to work. Any kind of help is appreciated.

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$\alpha,\beta,\gamma,\delta$ being the roots of the quartic, they verify the following

$$\begin{align} \alpha+\beta+\gamma+\delta&=-p\\ \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta&=q\\ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta&=-r\\ \alpha\beta\gamma\delta&=s \end{align}$$

Now denote $A=\alpha\beta+\gamma\delta$, $B=\beta\gamma+\alpha\delta$ and $\Gamma=\gamma\alpha+\beta\delta$. These three numbers are the roots of the following cubic.

$$X^3-(A+B+\Gamma)X^2+(AB+A\Gamma+B\Gamma)X-AB\Gamma=0$$

A straightforward computation gives

$$\begin{align} A+B+\Gamma&=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=q\\ AB+A\Gamma+B\Gamma&=pr-4s\\ AB\Gamma&=r^2-4qs+p^2s \end{align}$$

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  • $\begingroup$ Isn't there any elegant way? Something that involves less calculations? $\endgroup$ – SinTan1729 Nov 27 '16 at 11:15
  • $\begingroup$ I don't see any $\endgroup$ – marwalix Nov 27 '16 at 11:21
  • $\begingroup$ @SayantanSantra marwalix's answer has necessary algebraic calculation and is direct,... elegant. $\endgroup$ – Narasimham Nov 27 '16 at 11:29
  • $\begingroup$ Well, this kind of construction is trivial. What I was seeking is some method that gives the equation directly (i. e. by finding the value of a constant or something like that). But, since no one has posted such an answer, I'm marking it as the desired answer. $\endgroup$ – SinTan1729 Nov 27 '16 at 12:02

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