超急!!!maths~~factorization!!

x^6 + 2x^4y^2 + y^4x^2 ﹣4

= (x^4 + 2x^2y^2 + y^4)x^2 ﹣4


= x^2(x^2 + y^2)^2 ﹣4


= [(x^2 + y^2) x]^2 ﹣2^2


= [(x^2 + y^2) x ﹣2] [(x^2 + y^2)x + 2]


= (x^3 + xy^2 ﹣2) (x^3 + xy^2 + 2)

2008-03-03 18:12:15 補充：
利用恆等式: a^2 ﹣b^2 = (a ﹣b)(a + b)
利用恆等式: (a + b)^2 = a^2 + 2ab + b^2

2008-03-03 18:45:09 補充：
先抽出個x^6 + 2(x^4)(y^2) + (y^4)(x^2)中的因子x^2

然後x^4 + 2(x^2)(y^2) + y^4可利用恆等式: (a + b)^2 = a^2 + 2ab + b^2分解

設x^2為a，y^2為b，(x^2)^2 + 2(x^2)(y^2) + (y^2)^2

2008-03-03 18:45:15 補充：
再利用恆等式: a^2 ﹣b^2 = (a ﹣b)(a + b)分解[(x^2 + y^2) x]^2 ﹣2^2

設[(x^2 + y^2)x]為a，2為b，

最後可得出[(x^2 + y^2) x ﹣2] [(x^2 + y^2)x + 2]

可再化簡成(x^3 + xy^2 ﹣2) (x^3 + xy^2 + 2)

x^6 + 2(x^4)(y^2) + (y^4)(x^2) - 4
=(x^2)(x^4+2x^2y^2+y^4)-4
=(x^2)(x^2+y+2)^2-4
再仔細d,可以繼續
=(x(x^2+y+2))^2-2^2
=(x(x^2+y+2)+2)*(x(x^2+y+2)-2)

有d亂,唔知岩唔岩
自己check check

2008-03-03 18:36:43 補充：
最後打錯
(x(x^2+y+2)+2)*(x(x^2+y+2)-2)
正常應該係:
=(x(x^2+y^2)+2)*(x(x^2+y^2)-2)