(abc+bca+cba)(a-c+b)=

(abc+bca+cba)(a-c+b)=

應該是(abc+bca+cab)(a-c+b)
=111(a+b+c)(a+b-c)
=111[(a+b)²;-c²;]
=111a²;+222ab+111b²;-111c²;