By properties of determinants, we can split the given determinant into 2 parts

→

Taking x, y, z common from R₁, R₂, R₃ respectively

→

Operating R₁→R₁-R₃, R₂→R₂- R₃

→

→

Taking (x-z) and (y-z) common from R₁, R₂

→

Expanding with R₃

→y²+yz+z²-x²-xz-z² = xyz(xy²+xyz+xz²+zy²+yz²+z³-x²y-xyz-yz²-x²z-xz²-z³)

→ (y-x)(y+x) +z(y-x) =xyz(xy²+zy² -x²y -x²z)

→(y-x)(x+y+z)=xyz(xy(y-x)+z(y²-x²))

→(y-x)(x+y+z)= xyz(xy(y-x)+z(x+y)(y-x))

→(y-x)(x+y+z) = xyz(xy(y-x)+(xz+yz)(y-x))

→(y-x)(x+y+z)= xyz(y-x)(xy+xz+yz)

→x+y+z = xyz(xy+xz+yz)

Hence Proved