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. 
 A: 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 $$A=-\frac{2293235711}{49} + i\frac{2654208}{2401}.$$
A: Let $g(x)=b_3 x^3 + b_2 x^2 + b_1 x + b_0$ be a cubic having as zeros exactly $x_1^2+x_2^2$, $x_2^2+x_3^2$, $x_3^2+x_1^2$. Then $(x_1^2+x_2^2)(x_2^2+x_3^2)(x_3^2+x_1^2)=-b_0/b_3$.
Now $x_1^2+x_2^2=p_2-x_3^2$ etc. and so the roots of $g$ are given by $h(x_i)$ where $h(x)=p_2-x^2$. Note that $p_2=(x_1+x_2+x_3)^2-2(x_1 x_2 + x_2 x_3 + x_3 x_1)$ is easily computed from the coefficients of $f(x)=7x^3+24x^2+2016x+i$.
Finally, Wikipedia tells us that $g$ is given by the resultant $\operatorname {Res}_{x}(y-h(x),f(x))$, which you can find using WA.
