一道初三代数题已知x,y,z是正整数,且满足x^3-y^3-z^3=3xyz,x^2=2(y+z),求xy+yz+zx的值

一道初三代数题已知x,y,z是正整数,且满足x^3-y^3-z^3=3xyz,x^2=2(y+z),求xy+yz+zx的值
一道初三代数题
已知x,y,z是正整数,且满足x^3-y^3-z^3=3xyz,x^2=2(y+z),求xy+yz+zx的值

一道初三代数题已知x,y,z是正整数,且满足x^3-y^3-z^3=3xyz,x^2=2(y+z),求xy+yz+zx的值
x^3-y^3-z^3=3xyz
x^3-y^3-z^3-3xyz
=[(x-y)^3-3xy(x-y)]-z^3-3xyz
=[(x-y)^3-z^3]-3xy(x-y-z)
=[(x-y-z)^3-3(x-y)z(x-y-z)]-3xy(x-y-z)
=(x-y-z)^3-3(-xy+yz-zx)(x-y-z)
=(x-y-z)[(x-y-z)^2-3(-xy+yz-zx)]
=(x-y-z)(x^2+y^2+z^2+xy-yz+zx)=0
若x^2+y^2+z^2+xy-yz+zx=0
2x^2+2y^2+2z^2+2xy-2yz+2zx=0
(x^2+2xy+y^2)+(y^2-2yz+z^2)+(z^2+2zx+x^2)=0
所以(x+y)^2+(y-z)^2+(z+x)^2=0
所以只有x+y=0,y-z=0,z+x=0
所以y=z=-x
和正整数矛盾
所以x-y-z=0
x=y+z
x^2=2(y+z)
x^2=2x
x>0,x=2
y+z=2
则只有y=z=1
所以xy+yz+zx=2+1+2=5

分析：
由x^3-y^3-z^3=3xyz，得
x^3-y^3-z^3-3xyz=0
可联想到（x+y+z）（x2+y2+z2-xy-yz-zx）=x3+y3+z3-3xyz，

于是问题转化为（x-y-z）（x2+y2+z2+xy-yz-zx）=0的讨论，
再由x2=2（y+z）
即可求x，y，z的值．

x...

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分析：
由x^3-y^3-z^3=3xyz，得
x^3-y^3-z^3-3xyz=0
可联想到（x+y+z）（x2+y2+z2-xy-yz-zx）=x3+y3+z3-3xyz，

于是问题转化为（x-y-z）（x2+y2+z2+xy-yz-zx）=0的讨论，
再由x2=2（y+z）
即可求x，y，z的值．

x^3-y^3-z^3-3xyz
=x^3-（y+z）^3-3xyz+3y^2z+3yz^2
=（x-y-z）（x^2+xy+zx+y^2+2yz+z^2）-3yz（x-y-z）
=（x-y-z）（x^2+y^2+z^2+xy-yz+xz）=0
又∵x^2+y^2+z^2+xy-yz+zx
= [（x+y）^2+（y-z）^2+（z+x）^2]≠0．
故x-y-z=0，即x=y+z．
于是x^2=2x，所以x=2，y+z=2．
又∵x，y，z是正整数，y≥1，z≥1．
因此y=z=1，从而xy+yz+zx=5．

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由对称性可得：y=z
x^3-y^3-z^3=3xyz ==> x^3-2y^3=3xy^2 (1)
x^2=2(y+z) ==〉y=(1/4)x^2 (2)
(2) --> (1) 得:
x^3+6x^2-32=0 ==> x=2. y=z=1, xy+yz+zx=5

问题转化为（x-y-z）（x2+y2+z2+xy-yz-zx）=0的讨论，
再由x2=2（y+z）
即可求x，y，z的值
x^3-y^3-z^3=3xyz
x^3-y^3-z^3-3xyz
=[(x-y)^3-3xy(x-y)]-z^3-3xyz
=[(x-y)^3-z^3]-3xy(x-y-z)
=[(x-y-z)^3-3(x-y)z...

全部展开

问题转化为（x-y-z）（x2+y2+z2+xy-yz-zx）=0的讨论，
再由x2=2（y+z）
即可求x，y，z的值
x^3-y^3-z^3=3xyz
x^3-y^3-z^3-3xyz
=[(x-y)^3-3xy(x-y)]-z^3-3xyz
=[(x-y)^3-z^3]-3xy(x-y-z)
=[(x-y-z)^3-3(x-y)z(x-y-z)]-3xy(x-y-z)
=(x-y-z)^3-3(-xy+yz-zx)(x-y-z)
=(x-y-z)[(x-y-z)^2-3(-xy+yz-zx)]
=(x-y-z)(x^2+y^2+z^2+xy-yz+zx)=0
若x^2+y^2+z^2+xy-yz+zx=0
2x^2+2y^2+2z^2+2xy-2yz+2zx=0
(x^2+2xy+y^2)+(y^2-2yz+z^2)+(z^2+2zx+x^2)=0
所以(x+y)^2+(y-z)^2+(z+x)^2=0
所以只有x+y=0,y-z=0,z+x=0
所以y=z=-x
和正整数矛盾
所以x-y-z=0
x=y+z
x^2=2(y+z)
x^2=2x
x>0,x=2
y+z=2
则只有y=z=1
所以xy+yz+zx=2+1+2=5

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