Let $\Delta=\left|\begin{array}{ccc}\mathrm{x} & \mathrm{x}^2 & 1+\mathrm{px}^3 \\ \mathrm{y} & \mathrm{y}^2 & 1+\mathrm{py}^3 \\ \mathrm{z} & \mathrm{z}^2 & 1+\mathrm{pz}^3\end{array}\right|$, then
$\begin{aligned} \Delta &=\left|\begin{array}{lll}\mathrm{x} & \mathrm{x}^2 & 1 \\ \mathrm{y} & \mathrm{y}^2 & 1 \\ \mathrm{z} & \mathrm{z}^2 & 1\end{array}\right|+\left|\begin{array}{ccc}\mathrm{x} & \mathrm{x}^2 & \mathrm{px^{3 }} \\ \mathrm{y} & \mathrm{y}^2 & \mathrm{py}^3 \\ \mathrm{z} & \mathrm{z}^2 & \mathrm{pz} \mathrm{z}^3\end{array}\right| \\ &=\left|\begin{array}{lll}\mathrm{x} & \mathrm{x}^2 & 1 \\ \mathrm{y} & \mathrm{y}^2 & 1 \\ \mathrm{z} & \mathrm{z}^2 & 1\end{array}\right|+\mathrm{pxyz}\left|\begin{array}{ccc}1 & \mathrm{x} & \mathrm{x}^2 \\ 1 & \mathrm{y} & \mathrm{y}^2 \\ 1 & \mathrm{z} & \mathrm{z}^2\end{array}\right| \end{aligned}$
$=\left|\begin{array}{lll}1 & \mathrm{x} & \mathrm{x}^2 \\ 1 & \mathrm{y} & \mathrm{y}^2 \\ 1 & \mathrm{z} & \mathrm{z}^2\end{array}\right|+\operatorname{pxyz}\left|\begin{array}{ccc}1 & \mathrm{x} & \mathrm{x}^2 \\ 1 & \mathrm{y} & \mathrm{y}^2 \\ 1 & \mathrm{z} & \mathrm{z}^2\end{array}\right|=(1+\mathrm{pxyz})\left|\begin{array}{ccc}1 & \mathrm{x} & \mathrm{x}^2 \\ 1 & \mathrm{y} & \mathrm{y}^2 \\ 1 & \mathrm{z} & \mathrm{z}^2\end{array}\right|$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2, \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3$
$=(1+\text { pxyz })\left|\begin{array}{ccc}
0 & x-y & x^2-y^2 \\
0 & y-z & y^2-z^2 \\
1 & z & z^2
\end{array}\right|$
Taking common $(\mathrm{x}-\mathrm{y}) \&(\mathrm{y}-\mathrm{z})$ from $\mathrm{R}_1 \& \mathrm{R}_2$ respectively
$\begin{aligned}
&=(1+\text { pxyz })(x-y)(y-z)\left|\begin{array}{ccc}
0 & 1 & x+y \\
0 & 1 & y+z \\
1 & z & z^2
\end{array}\right| \\
&=(1+\text { pxyz })(x-y)(y-z)(z-x)
\end{aligned}$