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Determinant identity: (a + b + c)³

Short answer

Apply R1 → R1+R2+R3, factor (a+b+c), then C2−C1 and C3−C1 to obtain a triangular matrix. The determinant simplifies to (a+b+c)³.

Use properties of determinants to show that [a−b−c 2a 2a; 2b b−c−a 2b; 2c 2c c−a−b] (determinant) = (a + b + c)³.

Let s = a + b + c and

Δ = | a−b−c    2a      2a   |
    |  2b    b−c−a     2b   |
    |  2c      2c    c−a−b  |

Step 1: R₁ → R₁ + R₂ + R₃. The new first row is:

  • column 1: (a − b − c) + 2b + 2c = a + b + c
  • column 2: 2a + (b − c − a) + 2c = a + b + c
  • column 3: 2a + 2b + (c − a − b) = a + b + c

Take s common from R₁:

Δ = s | 1      1       1   |
      | 2b   b−c−a     2b  |
      | 2c     2c    c−a−b |

Step 2: C₂ → C₂ − C₁, C₃ → C₃ − C₁.

  • R₂: b − c − a − 2b = −(a + b + c) = −s, and 2b − 2b = 0
  • R₃: 2c − 2c = 0, and c − a − b − 2c = −s
Δ = s | 1   0    0  |
      | 2b −s    0  |
      | 2c  0   −s  |

The matrix is lower triangular: determinant = 1 · (−s) · (−s) = s².

Δ = s · s² = (a + b + c)³. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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