Zeroes of a polynomial. Evaluate an expression

Question

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.

Answer 1

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 $$A=-\frac{2293235711}{49} + i\frac{2654208}{2401}.$$

Answer 2

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.

Answer 3

If we define $f(x)=7(x-x_1)(x-x_2)(x-x_3) = 7 x^3 + 24 x^2 + 2016 x +\imath$ with $x_i$ the three distinct roots and consider the function $g(y)$ defined by $$ g(y)=(y-x_1^2)(y-x_2^2)(y-x_3^2) $$ then we find that the required quantity is given by $$ g(y=x_1^2+x_2^2+x_3^2) = (x_2^2+x_3^2)(x_1^2+x_3^2)(x_1^2+x_2^2) $$ Note that we also have $$ g(x^2)=(x-x_1)(x-x_2)(x-x_3)(x+x_1)(x+x_2)(x+x_3)=\frac{- f(x) f(-x)}{49} = \\x^6 + \frac{27648}{49} x^4 + (82944 - \frac{48}{49} \imath) x^2 + \frac{1}{49} $$ and hence that from the definition of $g(y)$ we know that $x_1^2+x_2^2+x_3^2 = -\frac{27649}{49}$, which gives the desired result $$ (x_1^2+x_2^2)(x_1^2+x_3^2)(x_2^2+x_3^2) = g(-\frac{27649}{49}) = -\frac{2293235711}{49} + \frac{1327104}{2401} \imath $$ which is indeed not a particular appealing result.

An alternative path is to follow the approach of @Eclipse Sun showed. This is based on introducing $s_1=x_1+x_2+x_3$, $s_2=x_1 x_2 + x_1 x_3 + x_2 x_3$, $s_3=x_1 x_2 x_3$ and realising that $$ f(x) = 7 (x^3 - s_1 x^2 + s_2 x - s_3) $$ You can check that: $$ (x_1^2+x_2^2)(x_1^2+x_3^2)(x_2^2+x_3^2) = s_1^2 s_2^2 - 2 s_2^2 - 2 s_1^3 s_3 + 4 s_1 s_2 s_3 - s_3^2 $$ and with the values $s_1=-24/7$, $s_2=2016/7$, and $s_3=-\imath/7$ you will find the same result as above.

Note that this is a little bit different from the solution of @Eclipse Sun, due to a small mistake $x_1^2 x_2^2 + x_1^2 x_3^2 + x_2^2 x_3^2 \neq \sigma_2^2 - \sigma_1 \sigma_3$ and which results in a wrong final answer.