n3 - 93 can be factorised into (n - 9)(n2 + 9n + 81)
N cubed
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
2 N cubed (3)
The GCF is 2mn.
3 cubed = 27 and 9 squared = 81 which is the biggest number
N cubed
n cubed = n times n times n
n squared x n n x n x n = n cubed n x n = n squared n squared x n = n cubed
'9 Cubed' means 9 X 9 X 9 9 x 9 = 81 )9 squared) Then 81 x 9 = 729 (9 cubed)
9 cubed is equal to 729.
n^3 (n cubed)
2 cubed is 4 6 cubed is 36 9 cubed is 81 & 8 cubed is 64
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.
9 cubed is equal to 729.
"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 '.