Zeroes of a polynomial. Evaluate an expression Let $x_1,x_2,x_3$ be the zeros of the polynomial $7x^3+24x^2+2016x+i$. Evaluate $(x_1^2+x_2^2)(x_2^2+x_3^2)(x_3^2+x_1^2)$.
My thoughts: I've tried $7(x-x_1)(x-x_2)(x-x_3)=0$ and expanded it out to match the polynomial given and got an ugly system of equations (which I can share). I'm not sure if I should start off with this equation or go a different way. 

The general method is to express the symmetric polynomial $A=(x_1^2+x_2^2)(x_2^2+x_3^2)(x_3^2+x_1^2)$ as a polynomial of elementary symmetric polynomials: $\sigma_1=x_1+x_2+x_3$, $\sigma_2=x_1x_2+x_2x_3+x_3x_1$, $\sigma_3=x_1x_2x_3$. Then you apply Vieta's formula.
Let's do the first step. Observe that
$$x_1^2+x_2^2+x_3^2=\sigma_1^2-2\sigma_2=s_2.$$
Thus, what you need is 
$$A=(s_2-x_1^2)(s_2-x_2^2)(s_2-x_3^2)=s_2^3-(x_1^2+x_2^2+x_3^2)s_2^2+(x_1^2x_2^2+x_1^2x_3^2+x_1^2x_3^2)s_2-x_1^2x_2^2x_3^2.$$
The first two terms are cancelled. Then we determine 
$$x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2=\sigma_2^2-2\sigma_1\sigma_3.$$
Hence, we get 
$$A=(\sigma_2^2-2\sigma_1\sigma_3)s_2-\sigma_3^2.$$
If you plug in all the numbers $\sigma_1=-24/7$, $\sigma_2=2016/7=288$, $\sigma_3=-i/7$, I believe it is a monster. WolframAlpha says it should be $$A=-\frac{2293235711}{49} + i\frac{2654208}{2401}.$$
