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Show that Δ = |((y + z)2, xy, zx), (xy, (x + z)2, yz), (xz, yz, (x + y)2)| = 2xyz(x + y + z)3

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Given that

Applying R1 →xR1, R2 → yR2 ,R3 → zR3 to Δ and dividing by xyz, we get

Taking common factors x, y, z from C1, C2 and C3 respectively, we get

Applying C2 → C2– C1, C3 → C3– C1, we have

Taking common factor (x + y + z) from C2 and C3, we have

Applying C2 → (C2 + 1/yC1) and C3 → C3 + 1/zC1, we get

Finally expanding along R1, we have 

Δ = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] 

= (x + y + z)2 (2yz) (x2 + xy + xz) 

= (x + y + z)3(2xyz)

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