Why is Excel not calculating the cube root as the cube root

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I think Excel treats exponents with recurring decimals written as fractions (like 1/3) differently than exponents without repeating decimals (like 1/2). It keeps me from using a formula that locates perfect cubes among a set of numbers.

Column A lists numbers 1 to 100.

Column B has the following formula (starting with row 5):

=IF($A5^(1/2)-ROUND($A5^(1/2),0)=0,1,0)

If the integer in column A is a perfect square, such as 1, 4, or 9, and does so correctly, this should return "1." The second formula I came up with, =IF(SQRT($A8)-ROUND(SQRT($A8),0)=0,1,0), also works.

Column C has the following formula (starting with row 5):

=IF($A5^(1/3)-ROUND($A5^(1/3),0)=0,1,0)

The only difference between this formula and the one for recognizing perfect squares is that there is a 3 where there was previously a 2. In the case of perfect cubes like 8, 27, 64, etc., this is not yielding a "1." (but does return a "1" for the number 1).
Can somebody correct this for me?

Nov 2 in Others by Kithuzzz
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