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

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

Using property 5,

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

$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&{{x^2}}&1\\y&{{y^2}}&1\\z&{{z^2}}&1\end{array}} \right| + pxyz\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\1&y&{{y^2}}\\1&z&{{z^2}}\end{array}} \right|$

Taking$x,y,z$ and p common from ${R_1},{R_2},{R_3}$

and ${C_3}$in determinant II.

Interchanging ${C_2} \leftrightarrow {C_3},$in determinant I.

$\Delta = - \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&1&{{x^2}}\\y&1&{{y^2}}\\z&1&{{z^2}}\end{array}} \right| + pxyz\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\1&y&{{y^2}}\\1&z&{{z^2}}\end{array}} \right|$

Interchanging ${C_1} \leftrightarrow {C_2},$

in determinant I.

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

$= (1 + pxyz)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\1&y&{{y^2}}\\1&z&{{z^2}}\end{array}} \right|$

Applying ${R_2} \to {R_2} - {R_1}$

and ${R_3} \to {R_3} - {R_1},$

we get
$\Delta = (1 + pxyz)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\0&{y - x}&{{y^2} - {x^2}}\\0&{z - x}&{{z^2} - {x^2}}\end{array}} \right|$

Taking $(y - x)$ and $(z - x)$ common from ${R_2}$

and ${R_3},$

we get
$\Delta = (1 + pxyz)(y - x)(z - x)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\0&1&{x + y}\\0&1&{z + x}\end{array}} \right|$

Applying ${R_2} \to {R_2} - {R_3},$

we get
$\Delta = (1 + pxyz)(y - x)(z - x)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&x&{{x^2}}\\0&0&{y - z}\\0&1&{z + x}\end{array}} \right|$

Expanding along ${C_1},$

we get
$\Delta = (1 + pxyz)(y - x)(z - x)\{ 0 - (y - z)\}$
$= (1 + pxyz)(x - y)(y - z)(z - x).$

Hence, proved.