分解因式X4+y4+z4-2x2y2-2y2z2-2x2z2,要详细过程，9：30之前，会做的告诉一声。
热心网友
x4+y4+z4-2x2y2-2y2z2-2x2z2=x4+y4+2x2y2-2y2z2-2x2z2+z4-4x2y2=(x2+y2)2-2z2(x2+y2)+z4-4x2y2=(x2+y2-z2)2-(2xy)2=(x2+y2-z2+2xy)(x2+y2-z2-2xy)=((x+y)2-z2)((x-y)2-z2)=(x+y+z)(x+y-z)(x-y+z)(x-y-z)
热心网友
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热心网友
对啊
热心网友
X4+y4+z4-2x2y2-2y2z2-2x2z2=(x^4+2x^2y^2+y^4)+z^4-2x^2z^2-2y^2z^2-4x^2y^2=(x^2+y^2)^2-2z^2(x^2+y^2)+z^4-4x^2y^2=(x^2+y^2-z^2)^2-4x^2y^2=[(x+y)^2-z^2][(x-y)^2-z^2]=(x+y-z)(x+y+z)(x-y-z)(x-y+z)
热心网友
你写得有点问题吧？X后面的2应该是表示平方吧？我把答案给你放在图片里，你看一下：
热心网友
X4+y4+z4-2x²y²-2x²y²-2y²z² =(x²+y²-z²)²-2x²y²+2x²y²+2y²z²-2x²y²-2x²y²-2y²z² =(x²+y²-z²)²-4x²y² =(x²+y²-z²+2xy)(x²+y²-z²-2xy) =[(x+y)²-z²][(x-y)²-z²] =(x+y-z)(x+y+z)(x-y-z)(x-y+z)
热心网友
=(x2-y2+z2)2-4x2z2=(x2-y2+z2)2-(2xz)2=(x2-y2+z2+2xz)(x2-y2+z2-2xz)=[(x+z)2-y2][(x-z)2-y2]=(x+z+y)(x+z-y)(x-z+y)(x-z-y)说明了就OK啦 我也有点过激 还是戾气太重...
热心网友
X4+y4+z4-2x2y2-2y2z2-2x2z2=(x^4+2x^2y^2+y^4)+z^4-2x^2z^2-2y^2z^2-4x^2y^2=(x^2+y^2)^2-2z^2(x^2+y^2)+z^4-4x^2y^2=(x^2+y^2-z^2)^2-4x^2y^2=(x^2+y^2-z^2+2xy)(x^2+y^2-z^2-2xy)=[(x+y)^2-z^2][(x-y)^2-z^2]=(x+y-z)(x+y+z)(x-y-z)(x-y+z)二楼的对不起了，刚才先提交了回来一看才发现没有写完整，所以补充完整请出题者采纳二楼的，我只是不想用一个不完整的答案放在上面