Determinant identity proof

Let

Δ = | x²  x²−(y−z)²  yz |
    | y²  y²−(z−x)²  zx |
    | z²  z²−(x−y)²  xy |

Step 1: C₂ → C₂ − C₁. The second column becomes −(y−z)², −(z−x)², −(x−y)². Take (−1) common:

Δ = −| x²  (y−z)²  yz |
     | y²  (z−x)²  zx |
     | z²  (x−y)²  xy |

Step 2: C₂ → C₂ + 2C₃. Since (y−z)² + 2yz = y² + z², etc.:

Δ = −| x²  y²+z²  yz |
     | y²  z²+x²  zx |
     | z²  x²+y²  xy |

Step 3: C₂ → C₂ + C₁. Every entry of the middle column becomes x² + y² + z². Take it common:

Δ = −(x²+y²+z²) | x²  1  yz |
                | y²  1  zx |
                | z²  1  xy |

Step 4: R₁ → R₁ − R₂, R₂ → R₂ − R₃.

• R₁: (x² − y², 0, yz − zx) = ((x−y)(x+y), 0, −z(x−y))
• R₂: (y² − z², 0, zx − xy) = ((y−z)(y+z), 0, −x(y−z))

Δ = −(x²+y²+z²) | (x−y)(x+y)   0   −z(x−y) |
                | (y−z)(y+z)   0   −x(y−z) |
                |     z²       1      xy    |

Step 5: Expand along C₂. The entry 1 is at (3, 2), sign (−1)⁵ = −1. Take (x−y) from R₁ and (y−z) from R₂:

inner = −(x−y)(y−z) | x+y  −z |
                     | y+z  −x |
      = −(x−y)(y−z) [ −x(x+y) + z(y+z) ]
      = −(x−y)(y−z) [ (z² − x²) + y(z − x) ]
      = −(x−y)(y−z)(z−x)(x+y+z)

Therefore Δ = −(x²+y²+z²) · [−(x−y)(y−z)(z−x)(x+y+z)]

Δ = (x − y)(y − z)(z − x)(x + y + z)(x² + y² + z²). ∎