【多項式】因式分解

[修‧虛無縹緲 10]

（ｘ＋ｙ＋ｚ）^５－ｘ^５－ｙ^５－ｚ^５

輪換對稱式？國中數學老師說解法麻煩？

摘自「奧林匹克競賽教程──初二分冊」

[Xiang 10]

令f(x,y,z) = (x+y+z)^5 - x^5 - y^5 - z^5

f(-y,y,z) = z^5 - (-y)^5 - y^5 - z^5 = 0

同樣的f(-z,y,z) = 0,f(x,-z,z) = 0

也就是當x + y = 0或x + z = 0或y + z = 0時,f(x,y,z) = 0

所以令f(x,y,z) = g(x,y,z)(x+y)(x+z)(y+z)

deg g = deg f - deg (x+y)(x+z)(y+z) = 2

且g(x,y,z) = g(y,x,z) = g(z,y,x) = g(x,z,y) = g(z,x,y) = g(y,z,x) = 0

所以令g(x,y,z) = a(x^2+y^2+z^2) + b(xy+xz+yz)

f(1,1,0) = (2a+b)*2 = 30

f(1,1,1) = (3a+3b)*8 = 240

(a,b) = (5,5)

則 f(x,y,z) = 5(x^2+y^2+z^2+xy+xz+yz)(x+y)(x+z)(y+z)

-

其實這對高中生來說都有點複雜...

不過既然是奧林匹克, 那倒頗正常QQ...

[曾阿牛 10]

大大提供的解答實在太帥了xd

[酷寶貝 10]

令f(x,y,z) = (x+y+z)^5 - x^5 - y^5 - z^5

f(-y,y,z) = z^5 - (-y)^5 - y^5 - z^5 = 0

同樣的f(-z,y,z) = 0,f(x,-z,z) = 0

也就是當x + y = 0或x + z = 0或y + z = 0時,f(x,y,z) = 0

所以令f(x,y,z) = g(x,y,z)(x+y)(x+z)(y+z)

deg g = deg f - deg (x+y)(x+z)(y+z) = 2

且g(x,y,z) = g(y,x,z) = g(z,y,x) = g(x,z,y) = g(z,x,y) = g(y,z,x) = 0

所以令g(x,y,z) = a(x^2+y^2+z^2) + b(xy+xz+yz)

f(1,1,0) = (2a+b)*2 = 30

f(1,1,1) = (3a+3b)*8 = 240

(a,b) = (5,5)

則 f(x,y,z) = 5(x^2+y^2+z^2+xy+xz+yz)(x+y)(x+z)(y+z)

-

其實這對高中生來說都有點複雜...

不過既然是奧林匹克, 那倒頗正常QQ...

為啥麼要這樣令???

不太了解???

[superman771214 10]

好酷的解法= =

[ck991021 10]

嗯～輪換對稱式的分解也是有對稱性的，

但是二次對稱式也不只這個(x^2+y^2+z^2)

我也想知道(x^2+y^2+z^2)是怎麼出來的(係數相乘？)