=[(x-y-z)-x][(x-y-z)^4+x(x-y-z)^3+x^2(x-y-z)^2+x^3(x-y-z)+x^4]-(y+z)(y^4-y^3z+y^2z^2-yz^3+z^4)
=-(y+z)[(x-y-z)^4+x(x-y-z)^3+x^2(x-y-z)^2+x^3(x-y-z)+x^4+y^4-y^3z+y^2z^2-yz^3+z^4]本回答被网友采纳
(x-y-z)^5-x^5-y^5-z^5因式分解
(x-y-z)^5-x^5-y^5-z^5 =[(x-y-z)-x][(x-y-z)^4+x(x-y-z)^3+x^2(x-y-z)^2+x^3(x-y-z)+x^4]-(y+z)(y^4-y^3z+y^2z^2-yz^3+z^4)=-(y+z)[(x-y-z)^4+x(x-y-z)^3+x^2(x-y-z)^2+x^3(x-y-z)+x^4+y^4-y^3z+y^2z^2-yz^3...
(x+y+z)^5-x^5-y^5-z^5
5 x z^4 + 5 y z^4 因式分解的话，结果是 5 (x + y) (x + z) (y + z) (x^2 + x y + y^2 + x z + y z + z^2)
分解因式(x+y+z)^5-x^5-y^5-z^5
(x+y+z)^5-x^5-y^5-z^5 =(x+y)(y+z)(z+x)[a(x²+y²+z²)+b(xy+yz+zx)]【其中，a和b是待定系数】代入x=y=1，z=0，得到 30=4a+2b ① 代入x=y=z=1，得到 240=24a+24b ② 解①②形成的方程组，得到 a=b=5 综上，(x+y+z)^5-x^5-y^5-z...
因式分解题请教 (x+y+z)^5-x^5-y^5-z^5 (请详细过程)
5*(x + y)*(x + z)*(y + z)*(x^2 + y*x + z*x + y^2 + z^2 + y*z)
(x+y+z)5-x5-y5-z5,5都为次方,作因式分解
因式分解=[（x+y+z)^5-x^5]-(y^5+z^5)=（x+y+z-x）[(x+y+z)^4+x(x+y+z)^3+ x^2(x+y+z)^2+x^3(x+y+z)+x^4]- (y+z)(y^4-y^3z+y^2z^2-yz^3+z^4)=(y+z)[(x+y+z)^4+x(x+y+z)^3+x^2(x+y+z)^2+ x^3(x+y+z)+x^4-y^4+y^3z-...
(x+y+z)5-x5-y5-z5做因式分解,5为次方
当X=-Y时,原式值为0,因此原式必有因式(X+Y).原式是对称式,同理也有因式(X+Z),(Y+Z).原式是五次式,而(X+Y)(Y+Z)(X+Z)是三次式,两者必然相差一个二次对称式：[K(X^2+Y^2+Z^2)+M(XY+YZ+XZ)]其中K,M为待定系数.即原式=(X...
分解因式请来答谢谢
（x+y+z)^5-x^5-y^5-z^5 =2(x+y)(y+z)(z+x)(x²+y²+z²+xy+yz+zx)
(x+y+z)^5-(x+y-z)^5-(x+z-y)^5-(z+y-x)^5,因式分解求高手解答
解 因为原式是5次齐次对称式 又由于x=0代入原式=0，知x是原式的因子，同理b,c也是原式的因子 所以可设 (x+y+z)^5-(x+y-z)^5-(x+z-y)^5-(z+y-x)^5=xyz[m(a²+b²+c²)+n(ab+bc+ca)]① x=y=z=1代入①得 m+n=80②;x=y=1,z=-1代入①得3m-...
(x-y)^5+(y-z)^5+(z-x)^5分解因式
回答：我想问,xyz大小关系
(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5
(x+y+z)^5-(x+y-z)^5-(x-y+z)^5-(-x+y+z)^5 解原式:=x-y-z^5-x-y+z^5-x+y-z^5+x-y-z^5 =(x-x-x-x)-(y-y-y-y)-(z^5-z^5-z^5-z^5)=0
