(2x^2+ax-y+6)-(-2bx^2-3x+5y-1)=(2+2b)x^2+(a+3)x-6y+7因为与字母x 所取的值无关那么(2+2b)=0,(a+3)=0那么b=-1,a=-31/3a^3-2b^2-(1/4a^3-3b^2)=1/3a^3-2b^2-1/4a^3+3b^2=1/12a^3+b^2当b=-1,a=-3时原式=-9/4+1=-5/4...