An algebra problem by CH Nikhil

Let $\alpha$ , $\beta$ and $\gamma$ be the roots of the cubic. Applying vieta's, we obtain $\begin{aligned}\alpha+\beta+\gamma&=-3p\\\alpha\beta+\beta\gamma+\gamma\alpha&=3q\\\alpha\beta\gamma&=-r\end{aligned}$
Since the roots are in A.P.
$\begin{aligned}2\beta&=\alpha+\gamma\\3\beta&=-3p\\\beta&=-p\\\beta\left(\alpha+\gamma\right)+\alpha\gamma&=3q\\2{\beta}^2-\frac{r}{\beta}&=3q\\2{p}^2+\frac{r}{p}&=3q\end{aligned}$ $\huge\color{#3D99F6}{\boxed{2{p}^3-3pq+r=0}}$

Let the three zeros in AP be $b-d$ , $b$ and $b+d$ , where $b$ is the middle term and $d$ is the common difference. By Vieta's formulas, we have:

$\begin{cases} 3p = - [(b-d)+b+(b+d)] = -3b & \Rightarrow b = -p \\ 3q = (b-d)b + b(b+d) + (b-d)(b+d) = 3b^2-d^2 & - d^2 = 3q - 3p^2 \\ r = -(b-d)b(b+d) = - b(b^2-d^2) = p(p^2 + 3q-3p^2) = 3pq - 2p^3 & \Rightarrow \boxed{2p^\color{#D61F06}{3}-3pq + r = 0} \end{cases}$

$\color{#D61F06}{\text{Note: The power should be 3 instead of 2}}$