Using properties of determinants, prove that:

$\Delta=\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|$

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

$\Delta=\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y-x & y^{2}-x^{2} & p\left(y^{3}-x^{3}\right) \\ z-x & z^{2}-x^{2} & p\left(z^{3}-x^{3}\right)\end{array}\right|$

$=(y-x)(z-x)\left|\begin{array}{ccc}x & x^{2} & 1+p x^{3} \\ 1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\ 1 & z+x & p\left(z^{2}+x^{2}+x z\right)\end{array}\right|$

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

$\Delta=(y-x)(z-x)\left|\begin{array}{llc}x & x^{2} & 1+p x^{3} \\ 1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\ 0 & z-y & p(z-y)(x+y+z)\end{array}\right|$

$=(y-x)(z-x)(z-y)\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ 1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\ 0 & 1 & p(x+y+z)\end{array}\right|$

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

$\begin{aligned} \Delta &=(x-y)(y-z)(z-x)\left[(-1)(p)\left(x y^{2}+x^{3}+x^{2} y\right)+1+p x^{3}+p(x+y+z)(x y)\right] \\ &=(x-y)(y-z)(z-x)\left[-p x y^{2}-p x^{3}-p x^{2} y+1+p x^{3}+p x^{2} y+p x y^{2}+p x y z\right] \\ &=(x-y)(y-z)(z-x)(1+p x y z) \end{aligned}$

Hence, the given result is proved.