已知x3+y3=27,x2y-xy2=6,求(y3-x3)+(x2y-3xy2)-2(y3-x2y)的值.
解:(y3-x3)+(x2y-3xy2)-2(y3-x2y)
=y3-x3+x2y-3xy2-2y3+2x2y
=-x3-y3+3x2y-3xy2.
因為x3+y3=27,所以-(x3+y3)=-27,即-x3-y3=-27,
因為x2y-xy2=6,所以3(x2y-xy2)=18,即3x2y-3xy2=18,
所以原式=-x3-y3+3x2y-3xy2=-27+18=-9.
∴(y3-x3)+(x2y-3xy2)-2(y3-x2y)的值為-9.
分析:本題考查整式的加法運算,由已知x3+y3=27,x2y-xy2=6,可以把(y3-x3)+(x2y-3xy2)-2(y3-x2y)轉化成已知項的形式代入求值.
點評:先對(y3-x3)+(x2y-3xy2)-2(y3-x2y)去括號合并同類項,轉化成含有x3+y3,x2y-xy2的形式.