Is there any shortcut to find if a number is a perfect cube? 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.
Example: $13689 =$ not ending with $2,3,7$ or $8$. We add up the digit $= 9$. So it's a perfect square $= 117$. But for $44$ does not end with $2,3,7,8$. We add up the digits $= 8$, hence 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 solutions. Just need a theoretical answer.
 A: I have been making something that needs to check a bunch of numbers and see if they are cubes or not.  This is what I found out in the process:


*

*If you take your number $n$ module $7$ and the result is not $0, 1$, or $6$ then it's not a cube.

*If $n$ module $13$ is not $0, 1, 5, 8$, or $12$ then $n$ is not a cube.

*There are many more like the first 2 examples which I have a list of at the bottom.

*If you know any of it's prime factors, like $2$ for example, then you can divide your number by $2^3$ (If it can't divide by $2^3$ then it's not a cube, but if it can then...) 
check if that is a cube.

*The best thing (that I know of) to do after those tests is to make guesses and then cube them and then pick higher or lower guesses and repeat.
EXAMPLE: $1718$.  Well maybe to start you guess $10$ which cubed is $1000$, since $1718 > 1000$ then we pick a higher number.  Say we pick $14$.  Well, $14^3 = 2744$ and since $1718 < 2744$ then next time we pick a lower guess... and repeat until we have it between $11$ and $12$ then since it is in between $1718$ isn't a cube.


LIST OF MODULE CHECKS:


*

*if $n$ module $7$ is not one of these: $0, 1, 6$ then $n$ is not a cube

*if $n$ module $9$ is not one of these: $0, 1, 8$ then $n$ is not a cube

*if $n$ module $13$ is not one of these: $1, 8, 12, 5, 0$ then $n$ is not a cube

*if $n$ module $19$ is not one of these: $1, 8, 7, 11, 18, 12, 0$ then $n$ is not a cube

*if $n$ module $31$ is not one of these: $1, 8, 27, 2, 30, 16, 29, 23, 4, 15, 0$ then $n$ is not a cube

*if $n$ module $37$ is not one of these: $1, 8, 27, 14, 31, 10, 26, 36, 6, 29, 23, 11, 0$ then $n$ is not a cube


Hope that gives you what you were looking for. :)
A: 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
