Δ = $\left|\begin{array}{ccc}\mathit{\alpha }& \mathit{\beta }& \mathit{\gamma }\\ {\mathit{\alpha }}^2& {\mathit{\beta }}^2& {\mathit{\gamma }}^2\\ \mathit{\beta }+\mathit{\gamma }& \mathit{\gamma }+\mathit{\alpha }& \mathit{\alpha }+\mathit{\beta }\end{array}\right|$
Applying ${\mathit{R}}_3$ → ${\mathit{R}}_3+{\mathit{R}}_1$
Δ =
= α + β + γ $\left|\begin{array}{ccc}\mathit{\alpha }& \mathit{\beta }& \mathit{\gamma }\\ {\mathit{\alpha }}^2& {\mathit{\beta }}^2& {\mathit{\gamma }}^2\\ 1& 1& 1\end{array}\right|$
Applying ${\mathit{C}}_1\to {\mathit{C}}_1-{\mathit{C}}_2$ and ${\mathit{C}}_2\to {\mathit{C}}_2-{\mathit{C}}_3$
Δ = α + β + γ $\left|\begin{array}{ccc}\mathit{\alpha }-\mathit{\beta }& \mathit{\beta }-\mathit{\gamma }& \mathit{\gamma }\\ {\mathit{\alpha }}^2-{\mathit{\beta }}^2& {\mathit{\beta }}^2-{\mathit{\gamma }}^2& {\mathit{\gamma }}^2\\ 0& 0& 1\end{array}\right|$
= α + β + γ (α - β) (β - γ) $\left|\begin{array}{ccc}1& 1& \mathit{\gamma }\\ \mathit{\alpha }\mathit{\beta }& \mathit{\beta }+\mathit{\gamma }& {\mathit{\gamma }}^2\\ 0& 0& 1\end{array}\right|$
= α+ β + γ (α - β) (β - γ) [1 (β + γ) - 1 (α + β)]
= (α - β) (β - γ) (α + β + γ) (+ γ - α - β)
= (α - β) (β - γ) (γ - α) (α + β + γ)
Hence proved.