Using properties of determinants, prove that:

$\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x),$ where $p$ is any scalar.

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

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

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

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

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

${\Delta  = (y - x)(z - x)\left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\ 
  1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)} \\ 
  0&{z - y}&{p(z - y)(x + y + z)} 
\end{array}} \right|}$

${ = (y - x)(z - x)(z - y)\left| {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + p{x^3}} \\ 
  1&{y + x}&{p\left( {{y^2} + {x^2} + xy} \right)} \\ 
  0&1&{p(x + y + z)} 
\end{array}} \right|}$

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

$\Delta=(x-y)(y-z)(z-x)\left[(-1)(p)\left(x y^{2}+x^{3}+x^{2} y\right)\right.$ $\left. { + 1 + p{x^3} + p(x + y + z)(xy)} \right]$

$ = (x - y)(y - z)(z - x)$ $\left[ { - px{y^2} - p{x^3} - p{x^2}y} \right.\left. { + 1 + p{x^3} + p{x^2}y + px{y^2} + pxyz} \right]$

$ = (x - y)(y - z)(z - x)(1 + pxyz)$

Hence, the given result is proved.