a^3m=3,b^3n=2
原式=a^(2m)^3+(b^n)^3-a^(2m)*b^n*a^4m*b^3n
=(a^3m)^2+b^(3n)-a^(6m)*b^(4n)
=3^2+2-[a^(3m)]^2*(b^3n)^(4/3)
=11-9×2^(4/3)
题目如果是：a^(2m)^3+(b^n)^3-a^(2m)*b^n*a^4m*b^2n
则原式=(a^3m)^2+b^(3n)-a^(6m)*b^(3n)
=9+2-9×2
=11-18
=-7