If $a+b+c=0, ab+bc+ca=1$ and $abc=1,$ then find the value of $\frac ab+\frac bc+\frac ca.$

Question

Clearly $a, b, c$ are the roots of the cubic equation: $x^3+x-1=0\tag{1}.$ We have to find: \begin{align} \frac ab+\frac bc+\frac ca&=\frac{a^2c+b^2a+c^2b}{abc}\\\\ &=a^2c+b^2a+c^2b\\\\ &=p,\text{ say}. \end{align} ($p$ is not a symmetric function of the roots.)

Now let: $q=ac^2+ba^2+cb^2.$ Then we have:

$0=\left(\sum ab\right)\left(\sum a\right)=p+s+3abc.$ This gives

$p+q=-3abc=-3.$

To find $p$ I multiplied $p$ and$q$ and obtained:

$pq=\sum a^3b^3+abc\left(\sum a^3\right)+3a^2b^2c^2.$

Now since $a, b, c$ are the roots of the Eq. $(1),$ so we can write:

$pq=\sum(1-a)(1-b)+abc\left[\sum(1-a)\right]+3(abc)^2=3-2\sum a+\sum ab+abc\left[3-\sum a\right]+3(abc)^2=3-0+1+1×(3-0)+3×1^2=10.$

This implies $p, q$ are the roots of the quadratic: $\color{green}{t^2+3t+10=0.}$ Which on solving gives

$\color{green}{t=\frac{-3\pm i\sqrt{31}}2\tag*{}.}$ Now my actual question is: between these two values of $t$ which one is $p$ and which one is $q$ ?

Please suggest. Thanks in advance.

Answer 1

Both, $p$ and $q$, are solutions. For instance if you change $a,b,c$ for $a,c,b$, the problem is the same, but $p$ goes to $q$.

Answer 2

$p$ and $q$ are not symmetric to the roots of the polynomial, but they are both conjugates

They form the quadratic $t^2+3t+10=0$, as it happens the root of the quadratic are the values of both $p$ and $q$, both of them satisfy the quadratic, which in turn satisfies symmetric function of the polynomial

I can therefore write that $p^2+3p+10=0$ and also $q^2+3q+10=0$, this means the value of $p$ is $\frac{-3+\sqrt{-31}}{2}$ or $\frac{-3-\sqrt{-31}}{2}$ and also $q$ is $\frac{-3+\sqrt{-31}}{2}$ or $\frac{-3-\sqrt{-31}}{2}$

So the change affected here just depends on the arrangement of the root we take that forms their structure $p = a^2c+b^2a+c^2b$ $q = a^2b+b^2c+c^2a$

So $p$ is any one of the solution of $t^2+3t+10=0$ while $q$ is the other