Take complex roots and apply it to a new polynomial Polynomial $f(x)=x^3-x^2+x+18$ has three distinct complex roots $r_1$,$r_2$, and $r_3$. Denote by $g(x)$ the cubic polynomial with leading coeffecient $1$ such that $g(r_i+\frac{1}{r_i})=0$, for $i=1,2,3$. The value of $g(2)$ can be expressed in the form $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute the sum of $m$ and $n$.
$\textbf{Thoughts}$
I'm not sure how to approach this question in any way whatsoever. Help is much appreciated.

Here is a different way through.
Let $y=x+\frac 1x$. Then $x^2=xy-1$. Substitute in the original equation to obtain $$(x-1)(xy-1)+x+18=0$$ or $$x^2y-xy+19=0=(xy-1)y-xy+19=x(y^2-y)-y+19$$ so that $$x=\frac {y-19}{y^2-y}$$
Then substitute back into $y=x+\frac 1x$ to obtain the cubic (it looks like quartic but the $y^4$ term cancels).

Note that once you have $x^2=xy-1$ it is clear that you can reduce the cubic in $x$ to a form which is linear in $x$.
