Determinant identity proof

Let D = |a b c; b c a; c a b| and let Δ be the determinant on the right-hand side:

Δ = | 2bc−a²   c²      b²   |
    |  c²    2ca−b²    a²   |
    |  b²      a²    2ab−c² |

Step 1: Evaluate D

Add C₂ and C₃ to C₁: every entry of C₁ becomes s = a + b + c.

D = s | 1  b  c |
      | 1  c  a |
      | 1  a  b |

R₂ → R₂ − R₁, R₃ → R₃ − R₁:

D = s | 1  b    c   |
      | 0  c−b  a−c |
      | 0  a−b  b−c |
  = s[(c−b)(b−c) − (a−c)(a−b)]
  = −s[(b−c)² + (a−c)(a−b)]
  = −s(a² + b² + c² − ab − bc − ca)
  = −(a³ + b³ + c³ − 3abc)

So D² = (a³ + b³ + c³ − 3abc)², which proves the first and last members are equal.

Step 2: Show Δ = D²

Let G = [−a −b −c; c a b; b c a]. Multiply D by G (row by column):

D·G = [a b c][−a −b −c]
      [b c a][ c  a  b]
      [c a b][ b  c  a]

• (1,1): −a² + bc + cb = 2bc − a²
• (1,2): −ab + ab + c² = c²
• (1,3): −ac + b² + ac = b²
• (2,1): −ab + c² + ab = c²
• (2,2): −b² + ac + ca = 2ca − b²
• (2,3): −bc + bc + a² = a²
• (3,1): −ac + ac + b² = b²
• (3,2): −bc + a² + bc = a²
• (3,3): −c² + ab + ba = 2ab − c²

So D·G is exactly the matrix whose determinant is Δ. Therefore Δ = |D|·|G|.

Step 3: |G| = D

Take (−1) out of R₁ of G: |G| = −|a b c; c a b; b c a|. The matrix [a b c; c a b; b c a] is D with R₂ and R₃ interchanged, so its determinant is −D. Hence |G| = −(−D) = D.

Conclusion

Δ = |D|·|G| = D · D = D² = (a³ + b³ + c³ − 3abc)²

Thus |a b c; b c a; c a b|² = Δ = (a³ + b³ + c³ − 3abc)². ∎