2+2 = 4 and 2*2 = 4. what is n if n+n+n = n*n*n?

Answer 1

3n=n^3.
Subtract 3n from both sides.
0=n^3-3n
Factor out an "n".
0=n(n^2-3)
Set each factor = 0.
n=0 and n^2-3=0
Solve for n.
n=0 and n= + or - square root of 3.

Answer 2

n=0

Answer 3

n=0

Answer 4

For n an integer you can show that there is no solution to the equation by induction.

We know it is false for n=1 and n=2

You need to prove that if it is false for any n>1, then it is false for n+1.

This can be done by showing 3n
Remove the brackets: 3n+3 < n^3 + 3n^2 + 3n + 1
Then 3n < n^3 + 3n^2 + 3n - 2
or 0 < n^3 + 3n^2 - 2
or 2 < n^3 + 3n^2

if n>1 then n^3 > 1 and 3n^2 > 1 so n^3 + 3n^2 > 2

This proves 3n < n^3 for all n >1.

Answer 5

n=√3&0

Answer 6

The trivial result is of course n = 0. There are no other integer solutions but there are two other real number solutions, which I will not reveal in case this is a homework problem. I will give you the hint that you may rewrite your question as looking for solutions to the equation x^3 - 3x = 0. Good luck.

Answer 7

The first answer dude got it right...
n + n + n = n ^ 3
3n = n^3
Divide both sides by n
3 = n^2
Therefore, n = +/- sqrt (3)

Also, the answer can be 0 since 3 * 0 = 0 ^ 3

Answer 8

Then 3n = n^3 => 3=n^2 => n= √3

Or n=0 (by inspection)


Doug

Answer 9

3 * n = n * n * n

n <> 0, then 3 = n * n, n = sqrt(3) or -sqrt(3)
you then verify if n= 0 is a solution. Yes, it is.

So the answer is:

n = 0, or sqrt(3) or -sqrt(3)

Answer 10

0

0+0+0=0
0*0*0=0

altenate answer

infinity. (works the same)