Several questions about working with Vieta's formulae on polynomials

Question

Let $f\in\mathbb{R}[X]$ be a polynomial of degree $3$ in the form of $f=ax^3+bx^2+cx+d.$ Find:

$E=\frac {x_1^2+x_2^2}{x_3^2}+\frac{x_2^2+x_3^2}{x_1^2}+\frac{x_1^2+x_3^2}{x_2^2}$.

Here I get stuck at finding the sum of squares taken $2$ by $2$. This can be rewritten as:

$E=(x_1^2+x_2^2+x_3^2)\frac{(x_1x_2)^2+(x_2x_3)^2+(x_1x_3)^2}{(x_1x_2x_3)^2}$

We can easily find: $x_1^2+x_2^2+x_3^2$ and $(x_1x_2x_3)^2$. But how to find the sum of the squared roots taken $2$ by $2$.

Also more questions:

How to find $\frac 1{x_1^2}+\frac 1{x_2^2}+\frac 1{x_3^2}$ and $\frac 1{x_1x_2}+\frac 1{x_2x_3} + \frac 1{x_1x_3}$?

Answer 1

Hint:

Add and subtract 3 to get $$E=\left(\sum x_i\right)^2\left(\frac {\sum (x_ix_j)^2}{(\prod x_i )^2}\right) -3$$

$$E=\left(\sum x_i\right)^2\left(\frac {\left(\sum x_ix_j\right)^2-2\prod x_i\left(\sum x_i\right) }{(\prod x_i )^2}\right) -3$$

Now substitute $\sum x_i=-b/a$, $\sum x_ix_j=c/a$,$\prod x_i=-d/a$

Answer 2

Hints: $x^3f(1/x)$ gives the polynomial with reciprocal roots and $f(\sqrt x)f(-\sqrt x)$ the one with squares of the original polynomial as roots... For calculating $E$ after your simplification, all you need is the latter and some Vieta.