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Is there any shortcut to find if a number is a perfect cube?

I am taking for instance finding if a number is a perfect square. So , if a number ends with 2,3,7,8. It cannot be a square. But if it does not end then it is not compulsory that the number is a perfect square. So we add up the digits to find if the sum is 1,4,7 or 9. If it is then it is a perfect square.

Ex: 13689 = not ending with 2,3,7 or 8. = We add up the digit = 9. So its a perfect square. = 117 . But for 44 = does not end with 2,3,7,8 = we add up the digits = 8 = not a perfect square.

So, i was wondering if there is any such rule to find if a number is a perfect cube.Please, no computing or algorithmic sols. . Just need a theoretical answer.

Thanks !

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    $\begingroup$ What exactly is digital cube root? What tools are the answerers allowed to use? For example does your programming tool kit include integer arithmetic for arbitrary size? Edit the question to make it unclear, please. Also include a description of what you have tried, and why it is not satisfactory (for example why "digital cube root" does not solve the problem). Otherwise the question may become a target of negative attention. Brevity is a virtue in some settings, but you appear to have overdone it to the point of making your question ununderstandable. $\endgroup$ – Jyrki Lahtonen Sep 22 '14 at 4:47
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    $\begingroup$ Well, if you lack access to a computer, calculator, or phone app, the easiest way by hand is to guestimate perfect cubes that are larger and smaller, then squeeze down until you have consecutive perfect cubes such that $n^3\leqslant x < (n+1)^3$. $\endgroup$ – Graham Kemp Sep 22 '14 at 4:48
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    $\begingroup$ There are easy-to-check necessary conditions that rule out many numbers: e.g., if it's even, it has to be divisible by $8$, etc. $\endgroup$ – user147263 Sep 22 '14 at 4:59
  • $\begingroup$ I am very sorry for this late reply. My Bad. I'll edit my post to explain my point. $\endgroup$ – Byteology Sep 23 '14 at 2:13
  • $\begingroup$ What you are asking for IS an algorithm (Look at the digits and do something with them). You could be more specific with your asking. (Looking for digit magic in particular). $\endgroup$ – amcalde Oct 20 '15 at 12:30
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This method will work for 1 followed by any number of zeros;

Let's take 1000 for example. 1000 has three zeros, and 3 is divisible by 3, so 1000 is a perfect cube Here: The prime factorisation of 1000 will give us 2*2*2*5*5*5. Since this is cube roots, we group the 2's and 5's to get 2*5=10, and 10 is the cube root of 1000.

Next example, 10000, which has 5 zeros. 5 can't be divided by 3 to get equal whole numbers, so 10000 is not going to be a perfect cube number. Here, the prime factorisation of 10000 will give us 2*2*2*2*2*5*5*13. We can't group the 5's and the 13, so this shows that 10000 is not a perfect cube number.

This method can work for only a 1 followed by how many zeros it has. Pretty much, if the number of zeros can be divisible by 3 to get equal whole numbers, it's a perfect cube number!

Something else you might want to know,

Take 89 cubed 89 cubed - 89*89*89 = 684909 The number of digits of the cube are equal to the number of 89's that are their, which is 3 89's. This shows that the number of digits that are there for the extended form of 89 cubed, ( which is 6 here) gives us how many digits there will be in the answer, which is 6. This is for odd numbers, but for even numbers it would change. Now let's take 34 34 cubed= 34*34*34. The extended form has 6 digits. But for this, we have to actually minus 1 form the total number of digits. 6-1=5. So, the answer should have 5 digits. Let's check. The cube of 34 gives us( drumroll please) 39304, which has ( drumroll) 5 digits! This method works for whole number cubing. Sorry if this is kind of complicated, and again if this does not satisfy you. By the way, I also had the same question, and did some research on the internet. I came up with nothing! The stuff I gave you above is pure observation and multiple testings! If you're not sure if they are not correct, try out those methods on some numbers and see if they are actually accurate ( I've already tried it, that's why I'm posting this here, duh)! Have fun!

-TheDRAGon

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