: I am having trouble with this could someone explain
and show me how to work this problem. Perform the 
operations. Write the ansers without negative
exponents. Assume that all varibles represent 
postive numbers. 
(27x^-3)^-1/3

(27x-3)-1/3

Give the 27 a 1 exponent

(271x-3)-1/3

Multiply the outer exponent -1/3 by each of the 
inner exponents:

271·-1/3x-3·-1/3

27-1/3x1

To get rid of the negative exponent, put the 
whole thing over 1,

 27-1/3x1
---------
   1

then move the base and exponent from numerator 
to denominator, and change the sign of 
the exponent to positive:

   x1
---------
 1·271/3

Erase the 1 exponent in the numerator and the
1 multiplier in the denominator:

   x
-------
 271/3

Now the 1/3 power is the same as the cube 
root, so now we have

   x
-------
 ³Ö27

The cube root of 27 is 3, so the final 
answer is

   x
  ---
   3

Edwin