Using properties of determinants, prove that: ∣∣∣∣∣x²+1xyxzxyy²+1yzxzyzz²+1∣∣∣∣∣=1+x²+y²+z²

LHS=∣∣∣∣∣x²+1xyxzxyy²+1yzxzyzz²+1∣∣∣∣∣
Taking out common factors x, y and z from R1, R2, R3 respectively, we get
=xyz∣∣∣∣∣∣x+1xyzxy+1yzxyz+1z∣∣∣∣∣∣
Applying R2→R2-R1
R3→R3-R1
=xyz∣∣∣∣∣∣x+1xyz𕒵x1y0𕒵x01z∣∣∣∣∣∣
Applying C1→xC1
C2→yC2
C3→zC3
=xyz × 1xyz∣∣∣∣x²+1y²z²��∣∣∣∣
=∣∣∣∣x²+1y²z²��∣∣∣∣
Expanding along R3, we have
LHS=𕒵∣∣∣y²z²10∣∣∣+1∣∣∣x²+1y²󔼓∣∣∣
=𕒵(-z²)+(x²+1+y²)
=1+x²+y²+z²
=RHS
Hence proved.