If $\alpha,\beta,\gamma,\delta$ are the roots of $x^4+px^3+qx^2+rx + s=0$, find in terms of $p,q,r,s$ the value of $\Sigma\frac{\alpha\beta}{\gamma }$

Question

If $\alpha,\beta,\gamma,\delta$ are the roots of $x^4+px^3+qx^2+rx + s=0$, find in terms of $p,q,r,s$ the value of $\Sigma\frac{\alpha\beta}{\gamma }$

My general strategy was transforming the equation to one whose roots are $\frac{\alpha\beta}{\gamma },etc$ but it seems to be impossible to narrow it down and presents a huge calculation.

Next, I tried simplifying the expression $\Sigma \frac{\alpha\beta}{\gamma }$ but, it doesn't turn out favourable as a huge calculation appears which couldn't be simplified more and no desirable form was obtained and this was in vain too. I don't understand how to approach it…

Answer 1

$x^4+px^3+qx^2+rx+s$
$=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$
$=x^4-(\alpha+\beta+\gamma+\delta)x^3+(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)x^2-(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta)x+\alpha\beta\gamma\delta$

So we have:
$p=-(\alpha+\beta+\gamma+\delta)$
$q=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta$
$r=-(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta)$
$s=\alpha\beta\gamma\delta$

This implies:

$$\sum\frac{\alpha\beta}{\gamma}=\frac{q(-r)-3s(-p)}{s}=3p-\frac{qr}{s}$$