f(x)=a(x^4-8x^2-9)+x^3+5x^2+6x

=a(x^2-9)(x^2+ 1)+x(x^2+5x+6)

=a(x+3)(x-3)(x^2+ 1)+x(x+2)(x+3)

=(x+3)[a(x-3)(x^2+ 1)+x(x+2)]，

So when x=-3, f(x) is always equal to 0, no matter whether A is an arbitrary number or not.

When x=3, whether a is an arbitrary number or not, f(x) is always equal to 90.

So there is always f(x) equal to 0, and there is always f(x) not equal to 0.