$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. $\alpha$, $\beta$, $\gamma$ 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... 

$$(x^3-1)^3=(x^2-px)^3$$
$$(x^3)^3-1-3x^3(x^3-1)=(x^3)^2-p^3(x^3)-3px^3(x^3-1)$$
Replace $x^3+1$ with $y$
$$(y-1)^3-3(y-1)(y-2)=(y-1)^2-p^3(y-1)-3p(y-1)(y-2)$$
$$y^3+(\cdots)y^2+(\cdots)y-1-6+1-p^3+6p=0$$
Now apply Vieta's formula
