证明:(x+y+z)^3xyz-(yz+zx+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+z^3x^3+x^3y^3)
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证明:(x+y+z)^3xyz-(yz+zx+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+z^3x^3+x^3y^3)
证明:(x+y+z)^3xyz-(yz+zx+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+z^3x^3+x^3y^3)
证明:(x+y+z)^3xyz-(yz+zx+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+z^3x^3+x^3y^3)
证明:∵(x+y+z)3xyz-(yz+zx+xy)3=xyz(x3+y3+z3)-(y3z3+z3x3+x3y3)
∴xyz[(x+y+z)3-(x3+y3+z3)]=(yz+zx+xy)3)-(y3z3+z3x3+x3y3)
∴xyz[(x3+y3+z3+3x2y+3xy2+3xz2 +3y2z +3yz2+6xyz)-(x3+y3+z3)],
=(y3z3+z3x3+x3y3+3y2z3x+3z3x2y+3y2zx2+3z2x3y+3zx3y2+6y2z2x2)-(y3z3+z3x3+x3y3),
∴xyz(3x2y+3xy2+3xz2 +3y2z +3yz2+6xyz)=3y2z3x+3z3x2y+3y2zx2+3z2x3y+3zx3y2+6y2z2x2
∴(3x3y2z+3x2y3z+3x2z3y+3y3z2x+3y2z3x+6x2y2z2=3y2z3x+3z3x2y+3y2zx2+3z2x3y+3zx3y2+6y2z2x2
∴(3x3y2z+3x2y3z+3x2z3y+3y3z2x+3y2z3x+6x2y2z2-3y2z3x-3z3x2y-3y2zx2-3z2x3y--6y2z2x2=0
∴(x+y+z)3xyz-(yz+zx+xy)3=xyz(x3+y3+z3)-(y3z3+z3x3+x3y3).
前面“战马1937”的证明,是由结论证明结论,没说清楚。应该是:
要证明:(x+y+z)^3xyz-(yz+zx+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+z^3x^3+x^3y^3) --- ①
即证明:(x+y+z)^3xyz-xyz(x^3+y^3+z^3)=(yz+zx+xy)^3-(y^3z^3+z^3x^3+x^3y^3) ---②
②左式=...
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前面“战马1937”的证明,是由结论证明结论,没说清楚。应该是:
要证明:(x+y+z)^3xyz-(yz+zx+xy)^3=xyz(x^3+y^3+z^3)-(y^3z^3+z^3x^3+x^3y^3) --- ①
即证明:(x+y+z)^3xyz-xyz(x^3+y^3+z^3)=(yz+zx+xy)^3-(y^3z^3+z^3x^3+x^3y^3) ---②
②左式=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:
②左式=3(a^2b+bc^2+a^2c+ab^2+b^2c+ac^2)
②右式=(a+b+c)^3-(a^3+b^3+c^3)=3(ab^2+ac^2+a^2b+b^2c+bc^2+a^2c)=左式
∴②式成立,∴原式成立。
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