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

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 $q$.
