如果代数式(2x^2+3ax-y+6)-(4bx^2-3x+5y-1)的值与字母x所取得值无关,试求代数式

(2x^2+3ax-y+6)-(4bx^2-3x+5y-1)
=2x^2+3ax-y+6-4bx^2+3x-5y+1
=2x^2-4bx^2+3ax+3x-y-5y+6+1
=（2-4b）x²+（3a+3）x-6y+7
与字母x所取得值无关
2-4b=0 3a+3=0
b=1/2 a=-1
3分之1a的立方-2b的平方-（4分之1a的立方-3b的平方）
=a³/3-2b²-a³/4+3b²
=a³/3-a³/4-2b²+3b²
=a³/12+b²
=（-1）³/12+（1/2）²
=-1/12+1/4
=1/6

(2x^2+3ax-y+6)-(4bx^2-3x+5y-1)
=(2-4b)x^2+(3a+3)x-(1+5)y+(6+1)
=(2-4b)x^2+(3a+3)x-6y+7

与字母x所取得值无关
2-4b=0
4b=2
b=0.5

3a+3=0
3a=-3
a=-1

3分之1a的立方...