2x^2+ax-y+6-2bx^2+3x+y-1
=(2-2b²)x+(a+3)x+5
值与字母的取值无关,则
2-2b²=0,a+3=0
解得b²=1,a=-3
1/3a^3-2b^2-1/4a^3+3b^2
=1/12a³+b²
=1/12×(-3)³+1
=-9/4+1
=-5/4