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