$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&{{x^2}}&{yz}\\y&{{y^2}}&{zx}\\z&{{z^2}}&{xy}\end{array}} \right| = (x - y)(y - z)(z - x)(xy + yz + zx)$

Let L.H.S. $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}x&{{x^2}}&{yz}\\y&{{y^2}}&{zx}\\z&{{z^2}}&{xy}\end{array}} \right|$

Multiplying ${R_1},{R_2}$

and ${R_3}$ by x, y and z respectively,

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

[Taking xyz common from ${C_3}$]
Applying ${C_2} \leftrightarrow {C_3},$

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

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

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

Applying ${R_1} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_1},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{x^2}}&{{x^3}}\\0&{{y^2} - {x^2}}&{{y^3} - {x^3}}\\0&{{z^2} - {x^2}}&{{z^3} - {x^3}}\end{array}} \right|$

Taking $(y - x)$ and $(z - x)$ common from ${R_2}$ and ${R_3}$respectively,

we get
$(y - x)(z - x)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{x^2}}&{{x^3}}\\0&{y + x}&{{y^2} + yx + {x^2}}\\0&{z + x}&{{z^2} + zx + {x^2}}\end{array}} \right|$

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

we get
$(y - x)(z - x)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{x^2}}&{{x^3}}\\0&{y - z}&{{y^2} - {z^2} + x(y - z)}\\0&{z + x}&{{z^2} + zx + {x^2}}\end{array}} \right|$

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

we get
$(y - x)(z - x)(y - z)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{{x^2}}&{{x^3}}\\0&1&{x + y + z}\\0&{z + x}&{{z^2} + zx + {x^2}}\end{array}} \right|$

Expanding along ${C_1},$

we get
$= \left( {y - x} \right)\left( {z - x} \right)\left( {y - z} \right)\left[ {\left( {{z^2} + zx + {x^2}} \right) - \left( {z + x} \right)\left( {x + y + z} \right)} \right]$

$= \left( {y - x} \right)\left( {z - x} \right)\left( {y - z} \right)\left[ {{z^2} + zx + {x^2} - zx - zy - {z^2} - {x^2} - xy - zx} \right]$

$= \left( {y - x} \right)\left( {z - x} \right)\left( {y - z} \right)\left[ { - xy - xz - zy} \right]$

$= \left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)\left( {xy + yz + zx} \right) = R.H.S.$
Hence, proved.