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)
