Determinant identity proof

Let

Δ = | a³   3a²     3a    1 |
    | a²  a²+2a   2a+1   1 |
    | a   2a+1    a+2    1 |
    | 1    3       3     1 |

Step 1: R₁ → R₁ − R₂, R₂ → R₂ − R₃, R₃ → R₃ − R₄.

• R₁: (a³ − a², 3a² − a² − 2a, 3a − 2a − 1, 0) = (a²(a−1), 2a(a−1), (a−1), 0)
• R₂: (a² − a, a² + 2a − 2a − 1, 2a + 1 − a − 2, 0) = (a(a−1), (a−1)(a+1), (a−1), 0)
• R₃: (a − 1, 2a + 1 − 3, a + 2 − 3, 0) = ((a−1), 2(a−1), (a−1), 0)

Step 2: take (a − 1) common from each of R₁, R₂, R₃ (factor (a − 1)³):

Δ = (a−1)³ | a²   2a    1   0 |
           | a   a+1    1   0 |
           | 1    2     1   0 |
           | 1    3     3   1 |

Step 3: expand along C₄, which has the single non-zero entry 1 at position (4, 4) (sign +):

Δ = (a−1)³ | a²  2a   1 |
           | a   a+1  1 |
           | 1   2    1 |

Step 4: R₁ → R₁ − R₃, R₂ → R₂ − R₃.

| a²−1   2a−2   0 |
| a−1    a−1    0 |
|  1      2     1 |

Expanding along C₃ (entry 1 at (3, 3)):

(a²−1)(a−1) − (2a−2)(a−1) = (a−1)²[(a+1) − 2] = (a−1)³

Therefore Δ = (a−1)³ · (a−1)³ = (a − 1)⁶. ∎