If the sum of two roots of the equation $x^3-p{x}^2+qx-r=0$ is zero, then 

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

that $\alpha +\beta =0$  Then,

$\alpha +\beta +\gamma =-\frac{(-p)}{1}\Rightarrow \alpha +\beta +\gamma =p\Rightarrow \gamma =p[\because \alpha +\beta =0]$

But, $\gamma$  is a root of the given equation. 

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\gamma }^3-p{\gamma }^2+q\gamma -r=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{p}^3-p^3+qp-r=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}pq=r\end{array} \left[\because \gamma =f_1\right.$