Given b^2(1/2b^-1)^2 (3b^3)^-2/(9b^4)-1
After looking over this question, I think it might write like this
b^2(1/2 b^-1)^2 (3b^3)^-2/(9b^4)^-1
In this problem we have a negative exponent to start with. That means we need to take the reciprocal of the base. Note that we DO NOT take the reciprocal of the exponent, only the base.
b^2 * (1/2b)^2 * (1/3b^3)^2 / (1/9b^4)
b^2 * (1/4b^2) * (1/9b^6) / (1/9b^4)
--- [remember when multiplying exponents to add them b^3 * b^3= b^6]
b^2/36b^8 / (1/9b^4)
simplify the numerator
1/36b^6 / ( 1/9b^4)
invert and multiply when dividing fractions.
------ * ------
[Typing it this way makes it easier to read, but can also be typed
1/36b^6 * 9b^4/1 ]
Of course this is assuming that there is a typo in the expression.
Hope this helps, and good luck.