Question
Using properties of determinants, prove the following:$\begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{vmatrix} = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)( \alpha + \beta + \gamma) $

Answer

$\text{LHS:} \triangle = \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta + \gamma & \gamma + \alpha & \alpha + \beta \end{vmatrix}$$\text{R}_{3} \rightarrow \text{R}_{3} + \text{R}_{1} = \triangle = \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \alpha +\beta+ \gamma &\alpha +\beta+ \gamma & \alpha +\beta+ \gamma \end{vmatrix}$
$= (\alpha + \beta + r) \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ 1 & 1 & 1 \end{vmatrix} $
$\text{Applying C}_{1} \rightarrow \text{C}_{1} -\text{C}_{2}$ $\text{and C}_{2} \rightarrow \text{C}_{2} - \text{C}_{3}$
$\triangle = ( \alpha + \beta + \gamma) \begin{vmatrix} \alpha- \beta & \beta - \gamma & \gamma \\ \alpha^{2} - \beta^{2} & \beta^{2} -\gamma^{2} & \gamma^{2} \\ 0 & 0 & 1 \end{vmatrix} $
Expanding by last row to get
$\triangle = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)( \alpha + \beta + \gamma) = \text{RHS} $

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