Factorize this question

a^6-b^6

=(a³)²-(b³)²-----------------先將它們拆開(a^6)=(a³)²≠a^5

=(a³+b³)(a³-b³)-------------抽(a³+b³)

=(a+b)(a²-ab+b²)(a³-b³)--------------利用恆等式:(a³+b³)≡(a+b)(a²-ab+b²)

=(a+b)(a²-ab+b²)(a-b)(a²+ab+b²)-------------利用恆等式:(a³+b³)≡(a-b)(a²+ab+b²)

驗算:

(a+b)(a²-ab+b²)(a-b)(a²+ab+b²)

=(a³+b³)(a³-b³)

=(a³)²-(b³)²

=a^6-b^6

所以這個答案是對的

1)a6-b6
=(a3)2-(b3)2
=(a3+b3)(a3-b3)
=(a+b)(a2-ab+b2)(a-b)(a2+ab+b2)


2007-11-06 21:07:57 補充：
恆等式係咪perfect square??