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Question

Prove that:
∣ ∣ ∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣ ∣ ∣=0

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Solution

We have to prove,
∣ ∣ ∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣ ∣ ∣=0
LHS =∣ ∣ ∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣ ∣ ∣=1xyz∣ ∣ ∣xy2z2xyzxy+xzx2yz2xyzyz+xyx2y2zxyzxz+yz∣ ∣ ∣
[R1xR1,R2yR2,R3zR3]
=1xyz(xyz)2∣ ∣yz1xy+xzxz1yz+xyxy1xz+yz∣ ∣
[taking (xyz) common from C1 and C2]
=xyz∣ ∣yz1xy+yz+zxxz1xy+yz+zxxy1xy+yz+zx∣ ∣[C3C3+C1]
=xyz(xy+yz+zx)∣ ∣yz11xz11xy11∣ ∣
[taking (xy+yz+zx) common from C3]
=0 [since, C2 and C3 are equal]
=RHS
Hence proved.


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