(2㎡+am-y+6)-(2b㎡-3m+5y-1)的值与m的值无关，求1/3a^3-2b^2-(1/4a^3-3b^2)的值

(2㎡+am-y+6)-(2b㎡-3m+5y-1)
=2㎡+am-y+6-2b㎡+3m-5y+1
=(2-2b)m^2+(a+3)m-6y+7
因为代数式的值m的值无关
所以它们的系数均为0
即：2-2b=0,a+3=0
b=1,a=-3
1/3a^3-2b^2-(1/4a^3-3b^2)
=1/3a^3-2b^2-1/4a^3+3b^2
=1/12a^3+b^2
代入得：原式=-9/4+1=-5/4

因为函数值与m无关
所以无论m取什么数函数值不变
所以原函数=(2m^2-2bm^2+am+3m)+(7-6y)
所以a=-3,b=1
所以1/3a^3-2b^2-(1/4a^3-3b^2)=-5/4

2m^2+am-y-6-2bm^2+3m-5y+1
=(2-2b)m^2+(a+3)m-6y-5
多项式的值与m的值无关，说明m的系数为0
所以2-2b=0,a+3=0
=>b=1,a=-3
1/3a^3-2b^2-(1/4a^3-3b^2)
=1/3a^3-2b^2-1/4a^3+3b^2
=1/12a^3+b^3
=-9/4+1
=-5/4