“计算(2x^3-3x^2y-xy^2)+(-x^3+2x^2y+y^3)-(x^3-x^2y-xy^2)的值,其中x=二分之一,

题目

“计算(2x^3-3x^2y-xy^2)+(-x^3+2x^2y+y^3)-(x^3-x^2y-xy^2)的值,其中x=二分之一,
之一“错抄成”x=-二分之一,但他计算结果也是正确的,试说明理由,并求出这个结果

答案

∵(2x^3-3x^2y-xy^2)+(-x^3+2x^2y+y^3)-(x^3-x^2y-xy^2)=2x³-3x²y-xy²-x³+2x²y+y³-x³+x²y+xy²=y³∵(2x^3-3x^2y-xy^2)+(-x^3+2x^2y+y^3)-(x^3-x^2y-xy^2)的值y
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