The condition that one root of $x^3+p{x}^2+qx+r=0$  is sum of the other two roots, is

$\alpha +\beta +\gamma =\frac{-b}{a}\left[\therefore \alpha +\beta =\gamma \right]$

$\gamma +\gamma =-p$

$2\gamma =-p$

$\gamma =\frac{-p}{2}$

Substitute in the equation

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

$\Rightarrow \frac{-p^3}{8}+\frac{p^3}{4}-\frac{pq}{2}+r=0$

$\Rightarrow \frac{-p^3}{8}\times 2+\frac{p^3}{4}\times 2-\frac{pq}{2}\times 2+r\left(2\right)=0$

$\Rightarrow \frac{-p^3}{4}+\frac{p^3}{2}-pq+2r=0$

$\Rightarrow \frac{-p^3+2p^3}{4}-pq+2r=0$

$\Rightarrow p^3=4\left(pq-2r\right)$