因式分解下列數式～3條

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1) x^6 +1
= (x^2)^3 + 1^3 = [(x^2) + 1][(x^2)^2 - (x^2) + 1^2] = (x^2 + 1)(x^4 - x^2 + 1) 2) y^6 -1
= (y^2)^3 - 1^3 = [(y^2) - 1][(y^2)^2 + (y^2) +1^2] = (y^2 - 1)(y^4 + y^2 + 1) 3) a^6-b^6
= (a^3)^2 - ( b^3)^2 = [a^3 - b^3][a^3 + b^3] = (a - b)(a^2 + ab + b^2)(a + b)( a^2 - ab + b^2)

第一題冇得分解
2)
y^6-1
=(y^3+1)(y^3-1)
3)
a^6-b^6
=(a^3+b^3)(a^3-b^3)

2011-09-07 22:31:39 補充：
(y^3+1)(y^3-1)
=y^3(y^3) +y^3x1 -y^3x1 +1^2
=y^6-1

(a^3+b^3)(a^3-b^3)
=a^3(a^3) + a^3b^3 -a^3b^3 +b^3(b^3)
=a^6-b^6

^that's the prove:)

2011-09-07 22:34:45 補充：
用了恆等式(identity)的方法

(a+b)(a-b)is identical to a^2-b^2

兩題也是喔