求证(x+y+z)^3xyz-(yz+xz+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)

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求证(x+y+z)^3xyz-(yz+xz+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)求证(x+y+z)^3xyz-(yz+xz+xy)^3=xyz(x

求证(x+y+z)^3xyz-(yz+xz+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)
求证(x+y+z)^3xyz-(yz+xz+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)

求证(x+y+z)^3xyz-(yz+xz+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)
证明:(x+y+z)^3xyz-xyz(x^3+y^3+z^3)=xyz((x+y+z)^3-(x^3+y^3+z^3))
=xyz(3xy^2+3xz^2+3x^2y+3Y^2z+3yz^2+3x^2z)
=3xyz(xy^2+xz^2+x^2y+y^2z+yz^2+x^2z)
令xy=a,yz=b,zx=c,并代入上式,得
(x+y+z)^3xyz-xyz(x^3+y^3+z^3)=3(a^2b+bc^2+a^2c+ab^2+b^2c+ac^2)①
又 令 xy=a,yz=b,zx=c,并代入 xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)
得 xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3) =(a+b+c)^3-(a^3+b^3+c^3)
=3(ab^2+ac^2+a^2b+b^2c+bc^2+a^2c)
=3a^2b+bc^2+a^2c+ab^2+b^2c+ac^2) ②
由①②得 (x+y+z)^3xyz-xyz(x^3+y^3+z^3)=xyz(x^3+y^3+z^3)-(y^3z^3+x^3z^3+x^3y^3)