若代数式(2x^2+ax-y+6)-(2bx^2-3x+5y-1)的值与x无关,求代数式3(a^2-2ab-b^2)-(

(2x^2+ax-y+6)-(2bx^2-3x+5y-1)
=(2-2b)x^2+(a+3)x-6y+7
所以
2-2b=0
a+3=0
即
b=1
a=-3
3(a^2-2ab-b^2)-(4a^2+ab+b^2)
=-a^2-7ab-4b^2
=-(-3)^2-7*(-3)*1-4*1^2
=8