From 2 identities and factoriz

6(a-1)b^3-(a-1)b^2
= (a-1)(6b^3-b^2)
= b^2(a-1)(6b-1)

2010-01-04 18:38:39 補充：
Any answer?
If you have, please tell me whether it's wrong or right.

2010-01-04 19:20:37 補充：
係6b^3-b^2 入面各term抽 b^2
所以變了 (a-1)*b^2*(6b-1)
即是b^2(a-1)(6b-1)

6(a-1)b^3-(a-1)b^2
=(a-1)(6b3-b2)
=b2(a-1)(6b-1)

2010-01-09 14:52:56 補充：
我把6同b3乘左(6b3),因為6b3同b2都要乘(a-1),so(6b3-b2),

2010-01-09 15:16:07 補充：
6b3-b2), 仲有得抽

2010-01-09 15:16:59 補充：
同(ab+bc)=b(a+c)道理一樣

2010-01-11 15:06:02 補充：
你可把b^2(6b-1)乘左去,你就會明白.

1) 6(a-1)b3-(a-1)b2
=6(a-1)(b2)(b)-(a-1)b2
=(a-1)(b2)(6b-1)