Given:

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

Let $\alpha , \beta \& \gamma$ are the roots of the given equation then

$\alpha +\beta +\gamma =-p ...\left(i\right)$

It is given that sum of any two roots is zero. So, let 

$\beta +\gamma =0$

Then, from $\left(i\right)$, we get

$\alpha =-p ...\left(ii\right)$

Since, $\alpha$ is also the root of the given equation therefore

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

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

$\Rightarrow pq=r$