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Determinant identity: 1

Short answer

Split the determinant into two: one equals abc times the Vandermonde determinant V, the other equals V. Thus Δ = (abc − 1)V. Since a,b,c are distinct, V ≠ 0, so Δ = 0 implies abc = 1.

If [a a² a³−1; b b² b³−1; c c² c³−1] (determinant) = 0, where a ≠ b ≠ c, apply the properties of determinants to show abc = 1.

Let Δ be the determinant:

Δ = | a  a²  a³−1 |
    | b  b²  b³−1 |
    | c  c²  c³−1 |

Step 1: Split the third column

Each entry of C₃ is a difference (a³ − 1), so by the additive property of determinants:

Δ = | a a² a³ |   −   | a a² 1 |
    | b b² b³ |       | b b² 1 |
    | c c² c³ |       | c c² 1 |

Step 2: Evaluate the first determinant

Take a, b, c common from R₁, R₂, R₃:

| a a² a³ |           | 1 a a² |
| b b² b³ | = abc  ·  | 1 b b² | = abc · V
| c c² c³ |           | 1 c c² |

where V is the Vandermonde determinant: V = (b − a)(c − a)(c − b).

Step 3: Evaluate the second determinant

Move C₃ to the first position: C₃ ↔ C₂, then C₂ ↔ C₁ (two interchanges, so the sign is unchanged):

| a a² 1 |   | 1 a a² |
| b b² 1 | = | 1 b b² | = V
| c c² 1 |   | 1 c c² |

Step 4: Combine

Δ = abc · V − V = (abc − 1) V = (abc − 1)(b − a)(c − a)(c − b)

Given Δ = 0 and a ≠ b ≠ c (all distinct, so V ≠ 0), we must have

abc − 1 = 0, i.e. abc = 1. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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