In regrads to the epression 9^2/3

Why is this not an equivilent statement to (3^2)^2/3

The reason i ask this question is because while working with exponents in two seperate problems it was handled two different ways.

In the problem 9^3/4 * 3^7/2 Over 27^3/2 and then continue on into the problem and it works fine

So why does it not work with the top problem

The book shows us chaging the 9 ^3/4 to (3^2)^3/4 and the 27^3/2 to (2^3)^3/2

Question

In regrads to the epression 9^2/3

Why is this not an equivilent statement to (3^2)^2/3

The reason i ask this question is because while working with exponents in two seperate problems it was handled two different ways.

In the problem 9^3/4 * 3^7/2 Over 27^3/2 and then continue on into the problem and it works fine

So why does it not work with the top problem

The book shows us chaging the 9 ^3/4  to (3^2)^3/4 and the 27^3/2 to (2^3)^3/2

anonymous · Answer

(9^2)/3 is equivalent to (3^2)^2/3 because (3^2)^2 is 9^2. I assume that is what you mean rather than 9^(2/3). 

The book is wrong when 27^3/2 is simplified to (2^3)^3/2. It should be (3^2)^3/2.

anonymous · Answer

Nevermind, thanks for the help i think i figured it out! all has to do with PEDAS Once you change 27 ^2/3 and make it (3^3)^2/3 you can not just add the expontes any longer like adding fractions because you must first deal with what is in the Parenthasis  if im understanding this correct
thats why i was confused

anonymous · Answer

What a second which brings me to my next question,