问几道初一得数学题，整式的题

3（a+b-c)+8(a-b-c)-7(a+b-c)-4(a-b-c)
=3a+3b-3c+8a-8b-8c-7a-7b+7c-4a+4b+4c
=-8b

3(a+b-c)+8(a-b-c)-7(a+b-c)-4(a-b-c)
=(3-7)*(a+b-c)+(8-4)*(a-b-c)
=-4(a+b-c)+4(a-b-c)
=-4a-4b+4c+4a-4b-4c
=-8b

解原式=3a+3b-3c+8a-8b-8c-7a-7b+7c-4a+4b+4c
=3a+8a-7a-4a+3b-8b-7b+4b-3c-8c+7c+4c
=(3+8-7-4)a+(3-8-7+4)b+(-3-8+7+4)c
=0-8b+0
=-8b
答：该式子的值为：-8b

解原式=3a+3b-3c+8a-8b-8c-7a-7b+7c-4a+4b+4c
=3a+8a-7a-4a+3b-8b-7b+4b-3c-8c+7c+4c
=(3+8-7-4)a+(3-8-7+4)b+(-3-8+7+4)c
=0-8b+0
=-8b

3（a+b-c)+8(a-b-c)-7(a+b-c)-4(a-b-c) ＝3a+3b-3c+8a-8b-8c-7a-7b+7c-4a+4b+4c=-8b

很简单啦。去括号嘛。
得：3A+3B-3C+8A-8B-8C-7A-7B+7C-4A+4B+4C=-8B