數學因式分解問題

16+2(a+b)^3
= 2(2)^3 + 2(a+b)^3
= 2[(2)^3 + (a+b)^3]
= 2[(2+(a+b))(2^2 –(2)(a+b) +(a+b)^2)]
= 2[(2+a+b)(4 -2a -2b +(a+b)^2)]
= 2(2+a+b)(4 -2a -2b +a^2+2ab +b^2)
= 2(2+a+b)( a^2 - 2a + b^2 - 2b + 2ab + 4)

(a+b)^3(a-b)^3
= [(a+b)(a-b)]^3
= (a^2 – b^2)^3


2012-10-04 01:38:02 補充：
Use these:
Factorize the sum of cubes
x^3 + y^3 = (x+y) (x^2 - xy + y^2)

Factorize difference of squares:
x^2 - y^2 = (x+y)(x-y)

(x^3)(y^3) = (xy)^3
Example:
(4^3)(5^3) = 20^3
(64)(125) = 8000
8000 = 8000

2012-10-05 23:59:58 補充：
I am afraid that the answer for question (1) given by 回答者：002 is wrong
16+2(a+b)^3 = 2(a+b+8)^3 ( ~ wrong)

Let us say a = 0, b = 0
16+2(a+b)^3 = 16 + 2(0+0)^3 = 16
2(a+b+8)^3 = 2(0 + 0 +8)^3 = 1024
16 is not equal to 1024

16+2(a+b)^3
=2[8+(a+b)^3]
=2(a+b+8)[64+8(a+b)+(a+b)^2]
=2(a+b+8)[(a+b)+8)^2
=2(a+b+8)(a+b+8)^2
=2(a+b+8)^3

(a+b)^3(a-b)^3
=[(a+b)(a-b)]^3