Question

If the roots of $x^3+3p{x}^2+3qx+r=0$ are in harmonical progression, show that $2q^3=r(3pq-r)$.

Answer 1

Given equation $x^3+3p{x}^2+3qx+r=0$

Let roots of the equation be $\alpha ,\beta ,\gamma$

Since the roots are in harmonic progression, $\frac{1}{\alpha }+\frac{1}{\gamma }=\frac{2}{\beta }$

$\Rightarrow \alpha \beta +\beta \gamma =2\alpha \gamma$ ....(1)

$\Rightarrow \alpha \beta +\beta \gamma +\alpha \gamma =3q$

$\Rightarrow 3\alpha \gamma =3q\left[\text{From (1)}\right]$

$\Rightarrow \alpha \gamma =q$ .....(2)

$\alpha \beta \gamma =-r⟹q\beta =-r⟹\beta =-\frac{r}{q}\left[\text{from (2)}\right]$

Since $\beta$ is a root of the given equation, therefore

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

$\Rightarrow {(-\frac{r}{q})}^3+3p{(-\frac{r}{q})}^2+3q(-\frac{r}{q})+r=0$

$\Rightarrow -r^2+3pqr-3q^3+q^3$

$\to 2q^3=3r(pq-r)$