Determinant identity: 2abc(a + b + c)³

Let

Δ = | (a+b)²   ca      bc   |
    |  ca    (b+c)²    ab   |
    |  bc      ab    (c+a)² |

Step 1: Remove the products

Multiply R₁ by c, R₂ by a, R₃ by b. Then

abc·Δ = | c(a+b)²   c²a      bc²   |
        |  ca²     a(b+c)²   a²b   |
        |  b²c      ab²    b(c+a)² |

Take c common from C₁, a common from C₂, b common from C₃. This gives a factor abc, which cancels with the abc on the left:

Δ = | (a+b)²   c²      c²    |
    |  a²    (b+c)²    a²    |
    |  b²      b²    (c+a)²  |

To keep the algebra simple put x = c, y = a, z = b, so that a + b = y + z, b + c = z + x and c + a = x + y:

Δ = | (y+z)²   x²      x²    |
    |  y²    (z+x)²    y²    |
    |  z²      z²    (x+y)²  |

Let s = x + y + z (= a + b + c).

Step 2: C₁ → C₁ − C₃, C₂ → C₂ − C₃

• Column 1: (y+z)² − x² = (y+z−x)s, 0, z² − (x+y)² = (z−x−y)s
• Column 2: 0, (z+x)² − y² = (z+x−y)s, z² − (x+y)² = (z−x−y)s

Take s from each of C₁, C₂ and write p = y+z−x, q = z+x−y, r = z−x−y:

Δ = s² | p  0  x² |
       | 0  q  y² |
       | r  r  (x+y)² |

Step 3: R₃ → R₃ − R₁ − R₂

• r − p = −2y, r − q = −2x, (x+y)² − x² − y² = 2xy

Δ = s² | p    0    x²  |
       | 0    q    y²  |
       | −2y −2x   2xy |

Step 4: Expand along R₁

Δ/s² = p(2xyq + 2xy²) + x²(2yq)
     = 2xy [ p(q + y) + xq ]

With q + y = z + x: p(z + x) + xq = (y+z−x)(z+x) + x(z+x−y) = yz + xy + z² − x² + xz + x² − xy = z(x + y + z) = zs.

So Δ = s² · 2xy · zs = 2xyz s³, i.e.

Δ = 2abc(a + b + c)³. ∎