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

Let Δ be the given determinant and s = a + b + c.

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

Step 1: C₁ → C₁ − C₃, C₂ → C₂ − C₃

Using x² − y² = (x − y)(x + y):

• Column 1: (b+c)² − a² = (b+c−a)s, 0, c² − (a+b)² = (c−a−b)s
• Column 2: 0, (c+a)² − b² = (c+a−b)s, c² − (a+b)² = (c−a−b)s

Take s out of each of C₁ and C₂ (factor s²). Write p = b+c−a, q = c+a−b, r = c−a−b:

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

Step 2: R₃ → R₃ − R₁ − R₂

• First entry: r − p = (c−a−b) − (b+c−a) = −2b
• Second entry: r − q = (c−a−b) − (c+a−b) = −2a
• Third entry: (a+b)² − a² − b² = 2ab

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

Step 3: Expand along R₁

Δ/s² = p(2abq + 2ab²) + a²(2bq)
     = 2ab[ p(q + b) + a q ]

Now q + b = c + a, so p(q + b) + aq = (b+c−a)(c+a) + a(c+a−b) = (bc + ab + c² − a²) + (ac + a² − ab) = c(a + b + c) = cs.

Therefore Δ/s² = 2ab · cs, i.e.

Δ = 2abc(a + b + c)³. ∎