Question

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

Answer 1

Δ=$\begin{vmatrix} \alpha &\alpha^2 &\beta+\gamma \\\ \beta&\beta^2 &\gamma+\alpha \\\ \gamma&\gamma^2 &\alpha+\beta \end{vmatrix}$
Applying R2$\rightarrow$R2-R1 and R3$\rightarrow$R3-R1,we have
=$\begin{vmatrix} \alpha &\alpha^2 &\beta+\gamma \\\ \beta+\alpha&\beta^2-\alpha^2 &\alpha+\beta \\\ \gamma-\alpha&\gamma^2-\alpha^2 &\alpha-\gamma \end{vmatrix}$

Applying R3$\rightarrow$R3-R2, we have:
Δ=(β-α)(γ-α)$\begin{vmatrix} \alpha &\alpha^2 &\beta+\gamma \\\ 1&\beta+\alpha &-1 \\\ 0&\gamma-\beta &0 \end{vmatrix}$

Expanding along R3,we have:
Δ=(β-α)(γ-α)[-(γ-β)(-α-β-γ)]
=(β-α)(γ-α)(γ-β)(α+β+γ)
=(β-γ)( γ-α)(α-β)(α+β+γ)

Hence,the given result is proved.