$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\\beta &{{\beta ^2}}&{\gamma + \alpha }\\\gamma &{{\gamma ^2}}&{\alpha + \beta }\end{array}} \right| = (\beta - \gamma )(\gamma - \alpha )(\alpha - \beta )(\alpha + \beta + \gamma )$

L.H.S.$= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &{{\alpha ^2}}&{\beta + \gamma }\\\beta &{{\beta ^2}}&{\gamma + \alpha }\\\gamma &{{\gamma ^2}}&{\alpha + \beta }\end{array}} \right|$

Applying ${C_3} \to {C_3} + {C_1},$

we get
L.H.S.$=$ $= \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &{{\alpha ^2}}&{\alpha + \beta + \gamma }\\\beta &{{\beta ^2}}&{\alpha + \beta + \gamma }\\\gamma &{{\gamma ^2}}&{\alpha + \beta + \gamma }\end{array}} \right|$

Taking ($\alpha + \beta + \gamma$) common from ${C_3}$,

we get
L.H.S. $=$ ($\alpha + \beta + \gamma$) $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &{{\alpha ^2}}&1\\\beta &{{\beta ^2}}&1\\\gamma &{{\gamma ^2}}&1\end{array}} \right|$

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

we get
L.H.S. $=$($\alpha + \beta + \gamma$)$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &{{\alpha ^2}}&1\\{\beta - \alpha }&{{\beta ^2} - {\alpha ^2}}&0\\{\gamma - \beta }&{{\gamma ^2} - {\beta ^2}}&0\end{array}} \right|$

Expanding along ${C_3}$,

we get
L.H.S. $=$ ($\alpha + \beta + \gamma$)$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{\beta - \alpha }&{{\beta ^2} - {\alpha ^2}}\\{\gamma - \beta }&{{\gamma ^2} - {\beta ^2}}\end{array}} \right|$

Taking $(\beta - \alpha )$ and $(\gamma - \beta )$

common from ${R_1}$ and ${R_2}$ respectively,

we get
$L.H.S. = (\alpha + \beta + \gamma )(\beta - \alpha )(\gamma - \beta )\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{\beta + \alpha }\\1&{\gamma + \beta }\end{array}} \right|$

$= (\alpha + \beta + \gamma )(\alpha - \beta )(\beta - \gamma )(\gamma - \alpha ) = R.H.S.$
Hence, proved.