Determinant identity: 2(ab + bc + ca)³

Let s = ab + bc + ca, σ = a + b + c, and let M be the matrix of the determinant

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

Step 1: Combine rows and columns with the matrix T

Let T = [−1 1 1; 1 −1 1; 1 1 −1]. Multiplying M by T on the left performs the row operations R₁ → R₂ + R₃ − R₁, R₂ → R₃ + R₁ − R₂, R₃ → R₁ + R₂ − R₃ (all applied together). Since |T| = 4, this multiplies the determinant by 4. M is symmetric, so doing the same combinations on the columns (multiplying by T on the right) multiplies it by another 4. Hence N = TMT satisfies |N| = 16Δ.

Using (b+c)² − c² − b² = 2bc and similar differences, the entries of N are

N = 2 [ 2a²+s    2ab−s    2ac−s ]
      [ 2ab−s    2b²+s    2bc−s ]
      [ 2ac−s    2bc−s    2c²+s ]

(for example N₁₁ = 2(2a² + ab + ac + bc), N₁₂ = 2(ab − ac − bc) = 2(2ab − s)).

Step 2: Evaluate the 3 × 3 determinant inside N

Let v = (a, b, c)ᵀ, e = (1, 1, 1)ᵀ, J = eeᵀ. The bracketed matrix equals

Q = s(2I − J) + 2vvᵀ

Put K = 2I − J. Its eigenvalues are 2 − 3 = −1, 2, 2, so |K| = −4, and K⁻¹ = ½(I − J) (check: (2I − J)(I − J) = 2I − 3J + J² = 2I, since J² = 3J).

By the matrix determinant lemma, |B + 2vvᵀ| = |B|(1 + 2vᵀB⁻¹v) with B = sK (taking s ≠ 0; the identity is polynomial, so it then holds for all a, b, c):

|B|      = s³ · (−4)
vᵀB⁻¹v   = (1/s)·½(vᵀv − (eᵀv)²) = (a²+b²+c² − σ²)/(2s) = −2s/(2s) = −1
|Q|      = −4s³ (1 + 2(−1)) = 4s³

Step 3: Combine

|N| = 2³ |Q| = 8 · 4s³ = 32 s³, so Δ = |N|/16 = 2s³.

Δ = 2(ab + bc + ca)³. ∎

(Spot check: a = b = c = 1 gives rows (4,1,1), (1,4,1), (1,1,4) with determinant 54 = 2 · 3³.)