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Determinant identity proof

Short answer

Apply column operations: C₂ → C₂ − C₁, then factor (b−c)² etc., add 2C₃, then C₂ + C₁ to get common factor. Result is (a−b)(b−c)(c−a)(a+b+c)(a²+b²+c²).

Find the value of the determinant [a² a²−(b−c)² bc; b² b²−(c−a)² ca; c² c²−(a−b)² ab].

Let

Δ = | a²  a²−(b−c)²  bc |
    | b²  b²−(c−a)²  ca |
    | c²  c²−(a−b)²  ab |

Step 1: C₂ → C₂ − C₁. The second column becomes −(b−c)², −(c−a)², −(a−b)². Take (−1) common:

Δ = −| a²  (b−c)²  bc |
     | b²  (c−a)²  ca |
     | c²  (a−b)²  ab |

Step 2: C₂ → C₂ + 2C₃. Since (b−c)² + 2bc = b² + c², and so on:

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

Step 3: C₂ → C₂ + C₁. All entries of the second column become a² + b² + c²; take it common:

Δ = −(a²+b²+c²) | a²  1  bc |
                | b²  1  ca |
                | c²  1  ab |

Step 4: R₁ → R₁ − R₂, R₂ → R₂ − R₃.

= −(a²+b²+c²) | (a−b)(a+b)   0  −c(a−b) |
              | (b−c)(b+c)   0  −a(b−c) |
              |     c²       1     ab    |

Step 5: Expand along C₂ (entry 1 at (3, 2), sign −1) and take (a−b) and (b−c) out:

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

because −a² − ab + bc + c² = (c² − a²) + b(c − a) = (c − a)(a + b + c).

Answer: Δ = (a − b)(b − c)(c − a)(a + b + c)(a² + b² + c²).

✓ Answer checked by Claude Opus 5.5 (high effort)

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