Answer: 1/b^2 or b^-2
By Laws of exponents we have the relationship of;
If m>n
a^m/a^n = a^m-n
If m<n
a^m/a^n = 1/a^n-m
And;
a^-n = 1/a^n
Also we have;
(a^m)*(a^n) = a^m+n
So let’s look at our problem;
b^-6/b^-4 = b^-6-(-4) = b^-2 = 1/b^2
Or we can do it this way;
b^-6/b^-4= (b^-6)*(b^4)= b^-6+4= b^-2= 1/b^2
If you have any questions please feel free to ask. Thanks and have a blessed day!