Pre - Rmo

We have abc=8 by Veita's. Since all must be integers a,b,c=+ or - {1,2,4,8}. Also a+b+c=-q by Vieta's and b-a=c-b => a+c=2b. Therefore a+b+c=3b=-q. Now abc=8 means we have (a,b,c)=(2,2,2) or (-4,-1,2) for a, b, and c to be in arithmetic progression. Since a,b,c are distinct, b=-1. q=-3b=-3*-1=3.

Let the three roots be $a$ , $b$ , and $c$ . By Vieta's formula , $a+b+c=-q$ . For arithmetic progression , $a+c=2b$ , $\implies 3b = - q$ . Since $b$ is a root of the equation,

$\begin{aligned} b^3+qb^2 - 2qb - 8 & = 0 & \small \color{#3D99F6} \text{Substituting }q = -3b \\ b^3-3b^3 + 6b^2 - 8 & = 0 \\ b^3 - 3b^2 + 4 & = 0 \\ (b+1)(b-2)^2 & = 0 \end{aligned}$

$\implies \begin{cases} b = -1 & \implies q = 3 & \implies x^3 + 3x^2 - 6x - 8 = 0 & \implies \color{#3D99F6} a = - 4, b = - 1, c = 2 & \small \color{#3D99F6} \text{Roots in AP} \\ b = 2 & \implies q = -6 & \implies x^3 - 6x^2 + 12x - 8 = 0 & \implies \color{#D61F06} b = 2 & \small \color{#D61F06} \text{Only one real root} \end{cases}$

Therefore the sum of all possible values of $p$ is $\boxed 3$ .

I think answer should be 3 as roots are different