Arithmetic Progression in Polynomials

Let the roots of $x^3-3px^2+qx-r=0$ be $a$ , $b$ and $c$ by Vieta's formula , we have:

$\begin{cases} a+b+c = 3p & ...(1) \\ ab+bc+ca = q & ...(2) \\ abc = r & ...(3) \end{cases}$

Since $a$ , $b$ and $c$ are in AP, then $a+c = 2b$ . Therefore, from $(1): \ a+b+c = 3b = 3p \implies b = p \quad ...(4)$ .

$\begin{aligned} (2): \quad ab+bc +ca & = q \\ (a+c)b + ca & = q \\ 2b^2 + ca & = q & \small \color{#3D99F6}{\text{Multiply both sides by }b} \\ 2\color{#3D99F6}{b}^3 + \color{#3D99F6}{abc} & = \color{#3D99F6}{b}q & \small \color{#3D99F6}{(3): \ abc = r, \ (4): \ b = p} \\ 2\color{#3D99F6}{p}^3 + \color{#3D99F6}{r} & = \color{#3D99F6}{p}q \end{aligned}$

$\implies \boxed{2p^3=pq-r}$

Let $\alpha-\beta,\alpha,\alpha+\beta\ \text{be zeroes of the given polynomial}\\ \alpha-\beta+\alpha+\alpha-\beta=3p\\ \alpha=p\\ \text{We know that,}\\ \alpha^{3}-3p\alpha^{2}+p\alpha-r=0\\ \text{Replacing}\ \alpha\ \text{by p,we get}\\ p^{3}-3p^{3}+pq-r=0\\ \boxed{2p^{3}=pq-r}$