Δ = |

α	β	γ
α2	β2	γ2
β+γ	γ+α	α+β

|
Applying R3 → R3+R1
Δ =
= α + β + γ |

α	β	γ
α2	β2	γ2
1	1	1

|
Applying C1→C1−C2 and C2→C2−C3
Δ = α + β + γ |

α−β	β−γ	γ
α2−β2	β2−γ2	γ2
0	0	1

|
= α + β + γ (α - β) (β - γ) |

1	1	γ
αβ	β+γ	γ2
0	0	1

|
= 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.