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.
You might start with $$g(x) = (x - (r_1 + 1/r_1))(x - (r_2 + 1/r_2))(x - (r_3 + 1/r_3))$$ EDIT:
So $$ \eqalign{g(2) &= \frac{(r_1 - 1)^2 (r_2 - 1)^2 (r_3 - 1)^2}{r_1 r_2 r_3} \cr &= \frac{f(1)^2}{r_1 r_2 r_3}}$$
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 $x$.
The brute-force approach would be to calculate the symmetric functions for $\,t_i = r_i+\frac{1}{r_i}\,$ using Vieta's relations, then determine the unique monic polynomial with roots $\,t_i\,$, and evaluate it at $2\,$.
The following is an alternative hint, noting that if $\,r\,$ is a root of $\,f\,$ then $\,r+\frac{1}{r}= 1 - \frac{18}{r^2}\,$.
Find first a polynomial $\,g(y)\,$ with roots $\,s_i=r_i^2\,$. For that, let $\,y=x^2\,$ and write the equation as $\,x(x^2+1) = x^2-18\,$, square and substitute $\,x^2=y\,$ to get $\,g(y) = y^3+y^2+37y-18^2\,$.
Then, a polynomial $\,h(z)\,$ with roots $\,t_i = 1 - \frac{18}{s_i}\,$ is $\,h(z)=(1-z)^3g\left(\frac{18}{1-z}\right)\,$ which after routine calculations gives $\,h(z)=-18(18 z^3 - 17 z^2 - 38 z + 361)\,$.
Therefore the monic polynomial with roots $\,r_i+\frac{1}{r_i}\,$ is $\,k(z)=\frac{-1}{18^2} h(z)\,$, and $\,k(2)\,$ follows.
