It may be complex

$x^3-2x^2+3x-4$

By Vieta's formula.
• $a+b+c=2$
• $ab+bc+ca=3$
• $abc=4$

$\Rightarrow \left(a+\frac{1}{a}\right)×\left(b+\frac{1}{b}\right)×\left(c+\frac{1}{c}\right)$

$=\frac{(abc)^2+\color{#20A900}{\left[(ab)^2+(bc)^2+(ca)^2\right]}+\color{#EC7300}{\left[a^2+b^2+c^2\right]}+1}{(abc)}$

$=\frac{(abc)^2+\color{#20A900}{\left[(ab+bc+ca)^2-2[(abc)(a+b+c)]\right]}+\color{#EC7300}{\left[(a+b+c)^2-2(ab+bc+ca)\right]}+1}{(abc)}$

$=\frac{(4)^2+\color{#20A900}{\left[(3)^2-2[(4)×(2)]\right]}+\color{#EC7300}{[(2)^2-2×(3)]}+1}{(4)}$

$=\frac{16-\color{#20A900}{7}-\color{#EC7300}{2}+1}{4}$

$=\frac{8}{4}$

$=\boxed{2}$

just put x=i and x=-i (i-a)(-i-a)=-(a²+1) Doing this for P(i)P(-i) = 8, and abc=4 (a²+1)(b²+1)(c²+1)/abc = 8/4 = 2