Factorisation

A^12 + B^12
= (A^4)^3 + (B^4)^3............用 a^3 + b^3 = (a+b)(a^2 - ab + b^2)
= (A^4 + B^4) (A^8 - A^4 B^4 + B^8)
= (A^4 + 2A^2 B^2 + B^4 - 2A^2 B^4) (A^8 + 2A^4 B^4 + B^8 - 3A^4 B^4)

作成平方差的形式,以下反覆運用 a^2 - b^2 = (a - b)(a + b)
= [ (A^2 + B^2)^2 - (√2AB)^2 ] [ (A^4 + B^4)^2 - (√3 A^2 B^2)^2 ]
= (A^2 + B^2 - √2AB)(A^2 + B^2 + √2AB) (A^4 + B^4 - √3 A^2 B^2)(A^4 + B^4 +√3 A^2 B^2)

= (A^2 + B^2 - √2AB)(A^2 + B^2 + √2AB) (A^4 + B^4 + 2A^2 B^2 -
(2+√3)A^2 B^2)(A^4 + B^4 + 2A^2 B^2 - (2-√3)A^2 B^2)

= (A^2 + B^2 - √2AB)(A^2 + B^2 + √2AB) {(A^2 + B^2)^2 - [√(2+√3) AB]^2}
{(A^2 + B^2)^2 - [√(2-√3) AB]^2}
= (A^2 + B^2 - √2AB)(A^2 + B^2 + √2AB) [A^2 + B^2 - √(2+√3)AB]
[A^2 + B^2 + √(2+√3)AB][A^2 + B^2 - √(2-√3) AB] [A^2 + B^2 +√(2-√3) AB]




2010-01-15 22:47:12 補充：
第四行 :
= (A^4 + 2A^2 B^2 + B^4 - 2A^2 B^2) (A^8 + 2A^4 B^4 + B^8 - 3A^4 B^4)
(不是B^4)

A^12+B^12
=(A^3+B^3)^4
=[(A+B)(A^2-AB+B^2)]^4
=(A+B)(A^2-AB+B^2)(A+B)(A^2-AB+B^2)(A+B)(A^2-AB+B^2)(A+B)(A^2-AB+B^2)

because:
there are two fomula:
#1:a^3+b^3=(a+b)(a^2-ab+b^2)
#2:a^-b^3=(a-b)(a^2+ab+b^2)

Answer is 5578