How do you simplify #.(a/b^3)^-3#?

1 Answer
Gió
Sep 22, 2015

Have a look:

Explanation:

You can consider that your negative cube is going to be applied to both numerator and denominator.
So you get:

#(a/b^3)^-3=((a)^-3)/((b^3)^-3)=#
you now use the rule of exponents where #(x^a)^b=x^(a*b)# to get:
#a^-3/(b^-9)#

At this point you can use (although I am not sure about the usefulness of it) the idea that a negative exponent sends your base to another "floor"; so if, say, #a# is at the numerator it can go downstairs to the denominator and viceversa.
In your case you can write:
#a^-3/(b^-9)=b^9/a^3#

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