Cubic equations and relationship between their roots If there exists a cubic equation $x^3 + 2x^2 +3x + 1 = 0$ has the roots $a,b,c$ 
And it is given that 


*

*$\frac{1}{a^3} + \frac{1}{b^3} - \frac{1}{c^3}$ 

*$\frac{1}{a^3} + \frac{1}{c^3} - \frac{1}{b^3}$ 

*$\frac{1}{c^3} + \frac{1}{b^3} - \frac{1}{a^3}$ 
Are the roots of another cubic 
$px^3 + qx^2 + rx + s$ 
And then what will the value of 
$p+q-12r+s$ 
be equal to? 
my work so far 
I initially tried establishing the relations of the sum of the roots taken 1,2 and 3 at a time and the coefficients that exists for any polynomial, I did this for the first and second equation but as I thought not much was simplified in the second equation which left me wondering if there was another shorter way, perhaps a trick or an observation that I am not able to make. I also tried expressing the cubic relation of $a,b, c$ in different ways by using the formulas for $(a+b+c)^3$ but that didn't result in much. I don't think I missed anything in the mentioned attempts that could be of significance but do feel free to check.
the answer, p+q-12r+s=5 
I wasn't even able to get close to an answer so. 
All help is greatly appreciated 

I have to sleep, so this is all I have right now. Perhaps I'll update tomorrow? It's too long for a comment anyways.
So, if we assume it's monic, obviously $p$ will be 1. Now, by Vieta's formulas, $-s$ will be the sum of the roots. And the sum of the roots turns out to be:
$\frac{1}{a^3} + \frac{1}{b^3} + \frac{1}{c^3}$
We want to express this in elementary symmetric polynomials, so I used Mathematica's SymmetricReduction function and got that it is equivalent to:
$$-\frac{3 a^2 b^2 c^2+(a b+a c+b c)^3-3 a b c (a+b+c) (a b+a c+b c)}{(abc)^3}$$
Use Vieta's formulas to get this equal to 12. So $s = 12$.
