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 $\Sigma\frac{\alpha\beta}{\gamma }$ 
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...
 A: $x^4+px^3+qx^2+rx+s$
$=(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$
$=x^4-(\alpha+\beta+\gamma+\delta)x^3+(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)x^2-(\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta)x+\alpha\beta\gamma\delta$
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:
$$\sum\frac{\alpha\beta}{\gamma}=\frac{q(-r)-3s(-p)}{s}=3p-\frac{qr}{s}$$
A: By Viète's formulas, the elementary symmetric polynomials of $\alpha,\beta,\gamma,\delta$ are:
$$e_1=\sum\alpha=-p,\quad e_2=\sum\alpha\beta=q,\quad e_3=\sum\alpha\beta\gamma=-r,\quad
e_4=\alpha\beta\gamma\delta=s.$$
Multiplying our sum of fractions by the common denominator $e_4,$ we get a symmetric polynomial:
$$e_4\sum\frac{\alpha\beta}\gamma=\sum\alpha\beta\gamma\delta\frac{\alpha\beta}\gamma=\sum\alpha^2\beta^2\delta=\sum\alpha^2\beta^2\gamma.$$
By the fundamental theorem on symmetric polynomials, $\sum\alpha^2\beta^2\gamma$ is a polynomial function of the $e_i$'s. To find out this (unique) function, apply one of the many algorithms. My favorite one is the one based on the lexicographic order on the (tuples of) exponents:

*

*Start from the exponents in $\alpha^2\beta^2\gamma:$ $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=(2,2,1,0).$

*Set $t_i:=\alpha_i-\alpha_{i+1}$ for $i<4:$ $t_1=0,t_2=1,t_3=1,$ and $t_4:=\alpha_4=0.$

*Compute $\prod e_i^{t_i}-\sum\alpha^2\beta^2\gamma=\sum\alpha\beta\sum\alpha\beta\gamma-\sum\alpha^2\beta^2\gamma=3\sum\alpha^2\beta\gamma\delta.$

*Iterate on this new symmetric polynomial (whose exponent is lexicographically  smaller), obtaining similarly: $\sum\alpha^2\beta\gamma\delta=e_1e_4.$
The final result is:
$$\sum\frac{\alpha\beta}\gamma=\frac{e_2e_3-3e_1e_4}{e_4}=\frac{-qr}s+3p.$$
