| 1 |
| a−b |
| a2 |
| a−c |
| b2 |
| b−c |
| c2 |
| (c−a)(c−b) |
=
| 1 |
| a−b |
| (a−b)(ab−ac−bc) |
| (a−c)(b−c) |
| c2 |
| (c−a)(c−b) |
=
| ab−ac−bc |
| (a−c)(b−c) |
| c2 |
| (c−a)(c−b) |
=
| ab−ac−bc+c2 |
| (a−c)(b−c) |
=1．
故答案为：1．
| a2 |
| (a−b)(a−c) |
| b2 |
| (b−c)(b−a) |
| c2 |
| (c−a)(c−b) |
| 1 |
| a−b |
| a2 |
| a−c |
| b2 |
| b−c |
| c2 |
| (c−a)(c−b) |
| 1 |
| a−b |
| (a−b)(ab−ac−bc) |
| (a−c)(b−c) |
| c2 |
| (c−a)(c−b) |
| ab−ac−bc |
| (a−c)(b−c) |
| c2 |
| (c−a)(c−b) |
| ab−ac−bc+c2 |
| (a−c)(b−c) |