Here's the question you clicked on:
yugobister
27^2/3
I'm assuming you mean 27^(2/3). If so, you first take the cube root of 27 (solve for the number which, cubed, yields 27). This number is 3. Then you square 3 to get 9. 9 is therefore the answer. Make sense?
kind of so if i do my next question: 64^(5/6). Then im not completely sure on what the next step would be.
The answer is 32. First, you have to find the number x that when taken to the sixth power yields 64. So x^6 should be 64.
\[27^{2/3}=\sqrt[3]{27^2}\]\[\sqrt[3]{27^2}=\sqrt[3]{729}\]\[\sqrt[3]{729}=9\]
In this case, that number is 2. Next you should take that x and take it to the fifth power (x^5) to find your final answer, which is 32.
\[64^{5/6}=\sqrt[5]{64^6}\]\[\sqrt[6]{64^5}=\sqrt[6]{1073741824}\]\[\sqrt[6]{1073741824}=32\]
i am still kind of confused. so for 1^5/3.
Again, you should first find out what number, when cubed, yields 1. That number here is simply 1. Then you should take that number (in this case, 1) and take it to the 5th power (1^5). So the answer is 1.
\[1^{5/3}=\sqrt[3]{1^5}\]\[\sqrt[3]{1^5}=\sqrt[3]{1}\]\[\sqrt[3]{1}=1\]