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.

(A) Let $\Delta = \left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ z & z^{2} & 1+p z^{3}\end{array}\right|$.
Using the property of determinants,we can split the third column:
$\Delta = \left|\begin{array}{lll}x & x^{2} & 1 \\ y & y^{2} & 1 \\ z & z^{2} & 1\end{array}\right| + \left|\begin{array}{lll}x & x^{2} & p x^{3} \\ y & y^{2} & p y^{3} \\ z & z^{2} & p z^{3}\end{array}\right|$
In the second determinant,take $p$ common from $C_3$,then $x, y, z$ common from $R_1, R_2, R_3$ respectively:
$\Delta = \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right| + pxyz \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|$
$\Delta = (1 + pxyz) \left|\begin{array}{lll}1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2}\end{array}\right|$
Applying $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$:
$\Delta = (1 + pxyz) \left|\begin{array}{lll}1 & x & x^{2} \\ 0 & y-x & y^{2}-x^{2} \\ 0 & z-x & z^{2}-x^{2}\end{array}\right|$
Taking $(y-x)$ common from $R_2$ and $(z-x)$ common from $R_3$:
$\Delta = (1 + pxyz)(y-x)(z-x) \left|\begin{array}{lll}1 & x & x^{2} \\ 0 & 1 & y+x \\ 0 & 1 & z+x\end{array}\right|$
Applying $R_3 \rightarrow R_3 - R_2$:
$\Delta = (1 + pxyz)(y-x)(z-x) \left|\begin{array}{lll}1 & x & x^{2} \\ 0 & 1 & y+x \\ 0 & 0 & z-y\end{array}\right|$
Expanding along $C_1$:
$\Delta = (1 + pxyz)(y-x)(z-x)(z-y)(1) = (1 + pxyz)(x-y)(y-z)(z-x)$.
Hence,the result is proved.