We have,
$$
x^3+p x^2+q x+r=0
$$
$\alpha, \beta, \gamma$ are roots of equation
$$
\begin{aligned}
\therefore \quad \alpha+\beta+\gamma & =-p \\
\alpha \beta+\beta \gamma+\alpha \gamma & =q \\
\alpha \beta \gamma & =-r
\end{aligned}
$$
Now, $(\alpha+\beta) \beta+\gamma)(\gamma+\alpha)$
$$
\begin{aligned}
& \left.=(\alpha+\beta) \beta \gamma+\alpha \beta+\alpha \gamma+\gamma^2\right) \\
& =\alpha \beta \gamma+\alpha^2 \beta+\alpha^2 \gamma+\alpha \gamma^2+\beta^2 \gamma+\alpha \beta^2+\alpha \beta \gamma+\beta \gamma^2 \\
& =2 \alpha \beta \gamma+\alpha^2 \beta+\alpha^2 \gamma+\alpha \gamma^2+\beta^2 \gamma+\alpha \beta^2+\beta \gamma^2 \\
& =(\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\alpha \gamma)-\alpha \beta \gamma \\
& =-p(q)-(-r)=r-p q
\end{aligned}
$$