Using properties of determinants, prove that:

$\Delta=\left|\begin{array}{lll}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have:

$\Delta=\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta-\alpha & \beta^{2}-\alpha^{2} & \alpha-\beta \\ \gamma-\alpha & \gamma^{2}-\alpha^{2} & \alpha-\gamma\end{array}\right|$

$=(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ 1 & \beta+\alpha & -1 \\ 1 & \gamma+\alpha & -1\end{array}\right|$

Applying $R_{3} \rightarrow R_{3}-R_{2}$, we have:

$\Delta=(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ 1 & \beta+\alpha & -1 \\ 0 & \gamma-\beta & 0\end{array}\right|$

Expanding along $R_{3}$, we have:

$\begin{aligned} \Delta &=(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)] \\ &=(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma) \\ &=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) \end{aligned}$

Hence, the given result is proved.