(1)若|a-1|+(b-2)2=0,A=3a2-6ab+b2,B=-a2-5,求A-B的值.
(2)试说明:无论x,y取何值时,代数式.
(x3+3x2y-5xy2+6y3)+(y3+2xy2+x2y-2x3)-(4x2y-x3-3xy2+7y3)的值是常数.
(1)∵A=3a2-6ab+b2,B=-a2-5,
∴A-B=(3a2-6ab+b2)-(-a2-5)=4a2-6ab+b2+5.
又∵|a-1|+(b-2)2=0,
∴A-B=4×12-6×1×2+22+5=1.
(2)原式=x3+3x2y-5xy2+6y3+y3+2xy2+x2y-2x3-4x2y+x3+3xy2-7y3,
=0.
原式化简值结果不含x,y字母,
∴无论x,y取何值,原式的值均为常数0.