![\sqrt[3]{\frac{27t^{9} }{729} }](/tpl/images/0870/5286/1c95f.png)
27 is a perfect cube for 
and 729 is a perfect cube for 
![\sqrt[3]{\frac{3^{3} t^{9} }{9^{3} } }](/tpl/images/0870/5286/0d074.png)
now you can pull out the numbers from under the root, for t (divide the exponent by 3)
now simply to What is the simplest form of 3^ √27t^9/729?
t^3/27
t^6/27
t^3/3
...
Mathematics, 05.11.2020 18:30 leeorareeves299
What is the simplest form of 3^ √27t^9/729?
t^3/27
t^6/27
t^3/3
t^6/3
Answers: 2
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