已知关于X的代数式(2X+AX-y+6)-(2bx^2-3x+5y-1)的值与x的取值无关,求式子1/3A^3-2B^2-1/4A^3+3B^2的值

(2X+AX-y+6)-(2bx^2-3x+5y-1)

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

∵值与X的取值无关
∴-2b=0 a+5=0
b=0 a=-5
1/3a^3-2b^2-1/4a^3+3b^2
=1/3a^3-1/4a^3+3b^2-2b^2
=1/12a^3+b^2
=1/12*(-5)^3+0^2
=1/12*(-125)

=-125/12

(2x+ax-y+6)-(2bx^2-3x+5y-1)
=2x+ax-y+6-2bx^2+3x-5y+1
=-2bx^2+2x+ax+3x-5y-y+6+1
=-2bx^2+(a+5)x-6y+7
代数式的值与x的取值无关
所以：-2b=0，a+5
1/3a^3-2b^2-1/4a^3+3b^2
=1/3a^3-1/4a^3+3b^2-2b^2
=1/12a^3+b^2
=1/12*(-5)^3+0^2
=1/12*(-125)
=-125/12

(2x+ax-y+6)-(2bx^2-3x+5y-1)
=2x+ax-y+6-2bx^2+3x-5y+1
=-2bx^2+2x+ax+3x-5y-y+6+1
=-2bx^2+(a+5)x-6y+7
代数式的值与x的取值无关
所以：-2b=0,a+5=0
即:b=0,a=-5
1/3a^3-2b^2-1/4a^3+3b^2
=1/3a^3-1/4a^3+3b^2-2b^2
=1/12a^3+b^2
=1/12*(-5)^3+0^2
=1/12*(-125)
=-125/12