FACTORIZATION既一條問題

(a+b)²c² - 4(a-b)(a+b)² + 4(a-b)²c²

=(a²+ab+ba+b²)+4(a-b)[(a-b)c²-(a+b)(a+b)]

=(a²+ab+ba+b²)+4(ab)[ac²-bc²-a²-2ab-b²]

Hope I can help you.

2008-11-26 18:32:24 補充：
Sorry the answer is
(a+b)²c² - 4(a-b)(a+b)² + 4(a-b)²c²

=(a²+ab+ba+b²)+4(a-b)[(a-b)c²-(a+b)(a+b)]

=(a²+ab+ba+b²)+4(ab)[ac²-bc²-a²-2ab-b²]

=c^2(3b-a)^2

(A+B)²C² - 4(A-B)(A+B)C² + 4(A-B)²C²

=c²[(A+B)² - 4(A-B)(A+B) + 4(A-B)²]............................抽公因式

=c²{(A+B)² - 2(2)(A-B)(A+B) + [2(A-B)]²}

=c²[(A+B)-2(A-B)]².............................................a²-2ab+b²=(a-b)²

=c²(A+B-2A+2B)²

=c²(3B-A)²

(a+b)^2 c^2-4(a-b)(a+b)c^2+4(a-b)^2 c^2
=c^2 [(a+b)-2(a-b)]^2......(a-b)^2=a^2-2ab+b^2
=c^2 (-a+3b)^2

(a+b)c - 4(a-b)(a+b) + 4(a-b)c
=(a+b)c - 2(a+b)2(a-b) + 4(a-b)c
={(a+b)2[c2 - 2]} {2(a-b)[(a-b)c2]}

= (a²+2ab+b²)c²-4(a-b)(a²+2ab+b²)+4(a²-2ab+b²)c²

= c²a²+2abc²+c²b²-4a^3-8a²b-4ab²-8ab²-4b^3+4a²c²-8abc²-4b²c²

=-4a^3-4b^3+5a²c²+5b²c²-6abc²-12a²b-12ab²

I hope this is it.><

可以用十字雙乘法計倒...
將
(a+b)c 分為 2個 (a+b)c
4(a-b)c 分為 2個 -2c(a-b)
可得出...
(a+b)c -2c(a-b)
(a+b)c -2c(a-b)
-------------------------------------
答案係 c^2 (3b-a)^2