# If zeroes of the polynomials f(x)=x^3-3px^2+qx-r are in AP then what is the relation between p, q and r?

Jan 13, 2018

$2 {p}^{3} + p q + r = 0$

#### Explanation:

Denoting the three zeros by $\alpha , \beta , \gamma$, we have:

$f \left(x\right) = {x}^{3} - 3 p {x}^{2} + q x - r$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} - \left(\alpha + \beta + \gamma\right) {x}^{2} + \left(\alpha \beta + \beta \gamma + \gamma \alpha\right) x - \alpha \beta \gamma$

In particular, equating the coefficient of ${x}^{2}$ we have:

$\alpha + \beta + \gamma = 3 p$

If $\alpha , \beta$ and $\gamma$ are in arithmetic progression with common difference $\delta$, then:

$\alpha = \beta - \delta$

$\gamma = \beta + \delta$

So:

$3 \beta = \left(\beta - \delta\right) + \beta + \left(\beta + \delta\right) = \alpha + \beta + \gamma = 3 p$

So:

$\beta = p$

That is: $p$ is one of the zeros of $f \left(x\right)$

So:

$0 = f \left(p\right) = \textcolor{b l u e}{{p}^{3}} - 3 p {\textcolor{b l u e}{p}}^{2} - q \textcolor{b l u e}{p} - r = - 2 {p}^{3} - p q - r$

Inverting the signs, that is:

$2 {p}^{3} + p q + r = 0$