Determinant identity proof

Short answer

Perform R₂→R₂+R₁ and factor a+b+c, then column operations and expansion along a row with zeros yield (a−b)(b−c)(a−c)(a+b+c).

Using properties of determinants, show that [a b c; b+c c+a a+b; a² b² c²] (determinant) = (a − b)(b − c)(a − c)(a + b + c).

Let

Δ = |  a    b    c  |
    | b+c  c+a  a+b |
    |  a²   b²   c² |

Step 1: R₂ → R₂ + R₁. Every entry of the new second row becomes a + b + c:

Δ = |  a    b    c  |
    | s    s    s  |      where s = a + b + c
    |  a²   b²   c² |

Take s common from R₂:

Δ = s | a   b   c  |
      | 1   1   1  |
      | a²  b²  c² |

Step 2: C₁ → C₁ − C₂, C₂ → C₂ − C₃.

Δ = s | a−b    b−c    c  |
      |  0      0     1  |
      | a²−b²  b²−c²  c² |

Step 3: Expand along R₂. The only non-zero entry is 1 in position (2, 3), with sign (−1)^(2+3) = −1:

Δ = −s [ (a−b)(b²−c²) − (b−c)(a²−b²) ]
  = −s (a−b)(b−c) [ (b+c) − (a+b) ]
  = −s (a−b)(b−c)(c−a)
  = s (a−b)(b−c)(a−c)

Therefore Δ = (a − b)(b − c)(a − c)(a + b + c). ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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