F.3 Factorization

1)(x + 4)3﹣x3
= [(x + 4)﹣x][(x + 4)2 + (x + 4)(x) + x2]
= (x + 4﹣x)(x2 + 8x + 16 + x2 + 4x + x2)
= 4(3x2 + 12x + 16)

2)x3﹣y3 + x﹣y
= (x﹣y)(x2 + xy + y2) + (x﹣y)
= (x﹣y)(x2 + xy + y2 + 1)

3) 8x3 + y3﹣2x﹣y
= (2x)3 + y3﹣(2x + y)
= (2x + y)[(2x)2﹣y(2x) + y2)﹣(2x + y)
= (2x + y)(4x2﹣2xy + y2)﹣(2x + y)
= (2x + y)(4x2﹣2xy + y2﹣1)

2008-10-14 18:07:04 補充：
By using the following idenities：
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)

2008-10-14 19:21:42 補充：
第1題，你可以當(x + 4)是a，而x是b
代入a^3 + b^3 = (a + b)(a^2 - ab + b^2)式中，
可得：[(x + 4)﹣x][(x + 4)^2 + (x + 4)(x) + x^2]

2008-10-14 19:21:57 補充：
其它的都是咁做。

2008-10-15 16:07:12 補充：
identity

1. = [(x + 4) - x][(x +4)^2 + x(x +4) + x^2]
= 4(x^2 + 8x + 16 + x^2 + 4x + x^2]
= 4(3x^2 + 12x + 16)
2.
= (x -y)(x^2 +xy + y^2) + (x - y)
=(x -y)(x^2 + xy + y^2 + 1)
3.
= (2x)^3 + y^3 - 2x - y
= [(2x) + y][(2x)^2 + y(2x) + y^2] - (2x +y)
= (2x +y)[4x^2 + 2xy + y^2 - 1]

2008-10-14 18:13:09 補充：
Correction: The last 2 lines should be -2xy, not +2xy, sorry for the mistake.