(2x²+ax-y+6)-(2bx²-3x-5y-1)
=(2-2b)x²+(a+3)x+4y+7
(2x²+ax-y+6)-(2bx²-3x-5y-1)的值与安母x的取值无关
则x²,x前的系数为0
2-2b=0 ;a+3=0
解得b=1 ,a=-3
(1/3)a³-2b³-[(1/4)a³-3b²]
=(1/12)a³+b³
=-27/12+1
=-5/4