Question

有这样一道题：“计算（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)的值

x=2009 y=1 甲同学你、把x=2009看成x=-2009。但结果仍正确。计算出最后的值，说说怎么回事？

Answer 1

（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)
=2x³-3x²y-2xy²-x³+2xy²-y³-x³+3x²y-y³
=-2y³
最后的值,与x无关

Answer 2

（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)
=2x³-3x²y-2xy²-x³+2xy²-y³-x³+3x²y-y³
=-2y³
最后的值,与x无关

Answer 3

将（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)化简会发现：
（2x^3-3x^2y-2xy^2)-(x^3-2xy^2+y^3)+(-x^3+3x^2y-y^3)=-2y^3
也就是说其值与x无关 so x抄错结果也正确

Answer 4

撒打算