Question

計算：

2c+a

(a-b)(b-c)(c-a)

+

b+c

(a-b)(c-b)(c-a)

-

b-a-c

(b-a)(c-b)(a-c)

．

Answer 1

原式=

2c+a

(a-b)(b-c)(c-a)

-

b+c

(a-b)(b-c)(c-a)

-

b-a-c

(a-b)(b-c)(c-a)

=

2c+a-b-c-b+a+c

(a-b)(b-c)(c-a)

=0．