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

Let s = a + b + c and

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

Step 1: R₁ → R₁ − R₃, R₂ → R₂ − R₃. Using x² − y² = (x − y)(x + y):

• R₁ − R₃ = ((b+c)² − a², 0, c² − (a+b)²) = ((b+c−a)s, 0, (c−a−b)s)
• R₂ − R₃ = (0, (c+a)² − b², c² − (a+b)²) = (0, (c+a−b)s, (c−a−b)s)

Take s common from R₁ and from R₂ (factor s²). Write p = b+c−a, q = c+a−b, r = c−a−b:

Δ = s² | p   0    r    |
       | 0   q    r    |
       | a²  b²  (a+b)² |

Step 2: C₃ → C₃ − C₁ − C₂.

• Row 1: r − p = (c−a−b) − (b+c−a) = −2b
• Row 2: r − q = (c−a−b) − (c+a−b) = −2a
• Row 3: (a+b)² − a² − b² = 2ab

Δ = s² | p   0   −2b |
       | 0   q   −2a |
       | a²  b²  2ab |

Step 3: Expand along R₁.

Δ/s² = p(2abq + 2ab²) + (−2b)(0·b² − q·a²)
     = 2ab·p(q + b) + 2a²b·q
     = 2ab [ p(q + b) + aq ]

With q + b = c + a: p(c + a) + aq = (b+c−a)(c+a) + a(c+a−b) = (bc + ab + c² − a²) + (ac + a² − ab) = c(a + b + c) = cs.

So Δ/s² = 2ab · cs, i.e.

Δ = 2abc(a + b + c)³. ∎

(This is the transpose of the determinant |(b+c)² a² a²; b² (c+a)² b²; c² c² (a+b)²|, which has the same value.)