Question 433233
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Do a prime factorization.


Start with 2.  81 is odd, so 2 is not a factor.


Consider 3.  The sum of the digits is divisible by 3, so 3 is a factor.


81 divided by 3 is 27


The sum of the digits of 27 is divisible by 3, so 3 is a factor.


27 divided by 3 is 9.


9 is divisible by 3


9 divided by 3 is 3.


Hence, the prime factorization of 81 is 3 times 3 times 3 times 3 times 3.


Since you are taking a cube root, group like factors in threes.  You have one group of three 3s.  Take three 3s out of the radical and leave one factor of 3 outside of the radical. One factor of 3 remains under the radical.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \sqrt[3]{81}\ =\ \sqrt[3]{3^3}\ \cdot\ \sqrt[3]{3}\ =\ 3\sqrt[3]{3}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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