don't know if I understand your question, but:
there are no numbers that are both cube numbers and prime numbers
8 is a factor of 16 that is a cube number (2^3)
2 is a factor of 16 that is a Prime number
Yes there are 6 numbers on a dice and the prime numbers are 2 3 and 5
If a factor appears 3 times, you get this factor (only once) times the cube root of a smaller number (the original number divided by the factor cubed).
It is the cube of a prime number.
Yes. For example 3 is the cube of cuberoot(3). However, by definition, a prime number cannot be a PERFECT cube.
All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
Yes there are 6 numbers on a dice and the prime numbers are 2 3 and 5
Three.Three.Three.Three.
If a factor appears 3 times, you get this factor (only once) times the cube root of a smaller number (the original number divided by the factor cubed).
5 and 8
It is the cube of a prime number.
Yes. For example 3 is the cube of cuberoot(3). However, by definition, a prime number cannot be a PERFECT cube.
The three prime numbers on the cube are: 2 3 and 5 so the probability is 3/6 or 1/2 simplified
343=7x7x7=73. 343 is a perfect cube. 7 is the prime factor of 343. 1,7, 49 and 343 are the distinct prime factors of 343.
All numbers have cube roots (not necessarily integral cube roots) so every prime has cube roots.
It doesn't matter if you know that the largest three-digit prime number is 997 or that its cube root is 9.98998998. If you include zero in the whole numbers, the answer will be zero.
343=7x7x7=73. 343 is a perfect cube. 7 is the prime factor of 343. 1,7, 49 and 343 are the distinct prime factors of 343.
No.