If $x_1, x_2, x_3, x_4$ are the roots of $x^4-x^3+2x^2+3x+1$, find $\frac{\sum x_i^3+\sum{(x_ix_jx_k)^3}}{\sum{(x_ix_j)^3}-(x_1x_2x_3x_4)^3}$ If $x_1, x_2, x_3, x_4$ are the roots of 
$x^4-x^3+2x^2+3x+1$
then find
$$\frac{\sum x_i^3+\sum{x_i^3x_j^3x_k^3}}{\sum{x_i^3x_j^3}-x_1^3x_2^3x_3^3x_4^3}$$
My attempt: I have tried many approaches but this attempt seems to have brought me nearest to a solution. First,$\sum{x_i^2}$ and $\sum{x_i^3}$ can be easily calculated since $\sum{x_i}, \sum{x_ix_j}, \sum{x_ix_jx_k}$ and $x_1x_2x_3x_4$ are already known. Then, $\sum{x_i^n}$ (n>3) can be calculated using the given biquadratic equation.
for example: $\sum{x_i^4}$ can be calculated by the equation:  $\sum{x_i^4}-\sum{x_i^3}+2\sum{x_i^2}+3\sum{x_i}+1=0$
Similarly, $\sum{x_i^5}$, $\sum{x_i^6}$,……..$\sum{x_i^n}$ can be calculated.
Now, using the Newton-Girard identity,
$\sum{(x_i^3)^4}$ -$\sum{(x_i^3)^3}$.$\sum{(x_i^3)}$ +$\sum{(x_i^3)^2}$.$\sum{x_i^3x_j^3}$ -$\sum{(x_i^3)}$.$\sum{x_i^3x_j^3x_k^3}$  +$x_1x_2x_3x_4$=0
I tried to find the relation between numerator and denominator terms.($\sum x_i^3, \sum{x_i^3x_j^3x_k^3}, \sum{x_i^3x_j^3}$ and $x_1^3x_2^3x_3^3x_4^3$).
 But nothing seems to work. Also, this attempt was way too lengthy. I am stuck up on this question for very long. First, I was looking for an elegant solution but now any solution will do. Please suggest a direction for solving this.

Hint: Let $P(x)=x^4-x^3+2x^2+3x+1$, and let $\omega\ne1$ be a third root of unity.  Then consider $Q(x^3)=P(x)P(\omega x)P(\omega^2 x)$.  Convince yourself that $Q(x)$ has roots $x_i^3$ and using Vieta you can read off the  terms you need from the coefficients of $Q$
Working that out, $Q(x)=x^4+14x^3+50x^2+6x+1$, so we need $\dfrac{-14-6}{50-1}=-\frac{20}{49}$.
