8c cubed
(a+b)(a squared-ab+b squared)
Of the natural numbers, the smallest (excluding 0 and 1 which are the same to any power) cubes are: 8 (2 cubed), 27 (3 cubed), 64 (4 cubed), and 125 (5 cubed).
x times x times x equals x cubed
64 should be between 27 and 125. (1 cubed equals 1. 2 cubed equals 8. 3 cubed equals 27. 4 cubed equals 64. Etc.)
N cubed
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
n3 - 93 can be factorised into (n - 9)(n2 + 9n + 81)
n^3 (n cubed)
2 N cubed (3)
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.
The GCF is 2mn.
. N times 6 cubed
"n cubed" is not an algebra problem, since it asks no question. It's an "expression", whose numerical value depends on the value of ' n '.
The statement n3 is ambiguous. I presume you mean n3, which means n cubed or n to the power of 3 (n*n*n). However, n3 (which should really be written as 3n) means n times 3 (n*3).
It is n cubed.
6n5