Yes. For example 3 is the cube of cuberoot(3). However, by definition, a prime number cannot be a PERFECT cube.
no its 5 cubed not 5 cubed 5
19
The sum of the first n cubed numbers is the square of the nth triangular number.
900=2x2x5x5x3x3 There are not any groups of three so the cube root is not a whole number. Now you have to take numbers and guess it. 10 cubed is 1000 9 cubed is 729 9.5 cubed is 902.5 So, keep trying to get close to the amount.
Numbers that have 2 factors are prime numbers; Numbers that have 4 factors are either: * the cube of a prime; or * the product of two different primes; Numbers that have 8 factors are either: * a prime to the power 7; * the product of a prime cubed and different prime; * the product of three different primes.
(3^3)*(11^3)=35937 so the cube root of 35937 using prime numbers is 3*11=33
Yes. For example 3 is the cube of cuberoot(3). However, by definition, a prime number cannot be a PERFECT cube.
no its 5 cubed not 5 cubed 5
19
The sum of the first n cubed numbers is the square of the nth triangular number.
900=2x2x5x5x3x3 There are not any groups of three so the cube root is not a whole number. Now you have to take numbers and guess it. 10 cubed is 1000 9 cubed is 729 9.5 cubed is 902.5 So, keep trying to get close to the amount.
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The sum of the first n cubed numbers is: [n*(n+1)/2]2 which is the same as the square of the sum of the first n numbers.
Numbers multiplied by it's self 3 times
2,628,071 and 2,628,073. 138 cubed is 2,628,072
Perfect cubes.