n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
6+N3 is the sum of N*3 then you add six. 6*N3 is you multiply 6 times three times N times three.
5+n+3 = 8+n
n3 + 2n2 - 15n = 0 : extract n as a factor of all the left hand terms then, n(n2 + 2n - 15) = 0 : the expression n2 + 2n - 15 can be factored as (n + 5)(n - 3) then, n(n + 5)(n - 3) = 0 : This holds true when either n = 0, or n + 5 = 0 or n - 3 = 0 When n + 5 = 0 then n = -5 : When n - 3 = 0 then n = 3. The solutions are n = 0 or n = -5 or n = 3.
no
n3 + 3n2 + 4n + 12 = (n3 + 3n2) + (4n + 12) = n2(n + 3) + 4(n + 3) = (n2 + 4)(n + 3).
Let any number be n:- n3/n3 = n*n*n/n*n*n = 1 And in index form: n3/n3 = n3-3 = n0 = 1
n3 + 1 = n3 + 13 = (n + 1)(n2 - n + 12) = (n + 1)(n2 - n + 1)
6+N3 is the sum of N*3 then you add six. 6*N3 is you multiply 6 times three times N times three.
5+n+3 = 8+n
n3 + 8n2 + 12n = n(n+2)(n+6)
Let the number be n.Then (n3)3 = n3 x 3 = n9.........or n to the power nine.However, if the question is what the the one-third power of a number cubed, then:(n3)1/3 = n3 x 1/3 = n1 = n
The formula for the nitride ion is N3-.
I assume that you mean n = 3 6n + 4 = 6(3) + 4 = 18 + 4 = 22
n3 + 2n2 - 15n = 0 : extract n as a factor of all the left hand terms then, n(n2 + 2n - 15) = 0 : the expression n2 + 2n - 15 can be factored as (n + 5)(n - 3) then, n(n + 5)(n - 3) = 0 : This holds true when either n = 0, or n + 5 = 0 or n - 3 = 0 When n + 5 = 0 then n = -5 : When n - 3 = 0 then n = 3. The solutions are n = 0 or n = -5 or n = 3.
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).
The ionic formula for Al3+ and N3- is AlN. Aluminum (Al) has a 3+ charge and nitrogen (N) has a 3- charge, so they combine in a 1:1 ratio to form a neutral compound.