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x3+y3+z3-3xyz 和 x3+y3+z3+3xyz分解因式
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人气:288 ℃ 时间:2020-04-01 07:53:46
解答
x^3+y^3+z^3-3xyz ==[( x+y)^3-3x^2y-3xy^2]+z^3-3xyz=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z) =(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z) =(x+y+z)(x^2+y^2+z^2-xy-xz-yz) 用...
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