$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. What is the product of all possible value of $p$?
Note that $p$ could be a complex number.
I tried some basic Vieta's formulas, couldn't find easy way to simplify…
Best Answer
$$x^3-x^2+px-1=0 \Rightarrow (x^3-1)^3-(x^2-px)^3=0 \Rightarrow x^9+(3p-.4)x^6+(p^3-3p+3)-1=0.$$ Let us transform this equation by $y=x^3+1 \rightarrow x=(y-1)^{1/3}$. Then we get a cubic Eq. for $y$ as $$y^3+y^2(3p-7)+(p^3-9p+14)y-p^3+6p-9=0,$$ $y_1, y_2, y_3$ are its roots. Then $$y_1 y_2 y_3=p^3-6p+9=2019 \Rightarrow p^3-6p-2010=0.$$ The roots of $p$ are $p_1,p_2,p_3$ and their product is: $p_1p_2p_3=2010$