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Question

Using properties of determinants, prove that
ω∣ ∣ ∣xyzx2y2z2y+zz+xx+y∣ ∣ ∣=(xy)(yz)(zx)(x+y+z)

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Solution

Let Δ=ω∣ ∣ ∣xyzx2y2z2y+zx+zx+y∣ ∣ ∣

Applying R3R3+R1, we get
Δ=ω∣ ∣ ∣xyzx2y2z2x+y+zx+y+zx+y+z∣ ∣ ∣
=(x+y+z)ω∣ ∣ ∣xyzx2y2z2111∣ ∣ ∣

Applying C1C1C2 and C2C2C3, we get
Δ=(x+y+z)ω∣ ∣ ∣xyyzzx2y2y2z2z2001∣ ∣ ∣
=(x+y+z)(xy)(yz)ω∣ ∣11zx+yy+zz2001∣ ∣

Expanding along R3, we get
=(x+y+z)(xy)(yz)(y+zxy)
=(xy)(yz)(zx)(x+y+z)

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