What's the solution to POTW #476?

[anemone]

Here is this week's POTW:

-----Let $a,\,b,\,c$ and $d$ be the roots to the polynomial $f(x)=x^4-3x^3+2x^2+5x-4$ . Evaluate $\left(a+1+\dfrac{1}{a}\right)\left(b+1+\dfrac{1}{b}\right)\left(c+1+\dfrac{1}{c}\right)\left(d+1+\dfrac{1}{d}\right)$.
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[anemone]

No one answered last two week's POTW. However, you can read the suggested solution as follows:

Spoiler

Let $w=e^{\frac{2\pi i}{3}}$ so

$\begin{align*}P(x)&=x^4-3x^3+2x^2+5x-4\\&=2w^2+6w-7\\&=2\left(-\dfrac{1}{2}-\dfrac{i\sqrt{3}}{2}\right)+6\left(-\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right)-7\\&=-1+i\sqrt{3}-3+3i\sqrt{3}-7=2i\sqrt{3}-11\end{align*}$

With this we see that what we are intended to find is indeed $\displaystyle \prod_{f(r)=0}\dfrac{r^2+r+1}{r}=(-1)^4\prod_{f(r)=0}\dfrac{(w-r)(w^2-r)}{(0-r)}=\dfrac{P(w)P(w^2)}{P(0)}$.

Since $w^2=\overline{w}$, $P(w^2)=\overline{P(w)}$, so $P(w^2)=-2i\sqrt{3}-11$ and so

$P(w)P(w^2)=|2i\sqrt{3}-11|^2=121+12=133$

and $P(0)=-4$,

Hence $\left(a+1+\dfrac{1}{a}\right)\left(b+1+\dfrac{1}{b}\right)\left(c+1+\dfrac{1}{c}\right)\left(d+1+\dfrac{1}{d}\right)=-\dfrac{133}{4}$.