Mathematical induction prove that 9 divides $n^3 + (n+1)^3 + (n+2)^3$ .

Question

How can I use mathematical induction to prove that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a nonnegative integer?

Answer 1

By noting that $$(n+3)^3=n^3+9n^2+27n+27=n^3+9(n^2+3n+3)$$ for any $n$. That will let you complete the induction step. The base step is easy.

Answer 2

The same way you usually use induction. First check $n=1$. We have $1^3+2^3+3^3=36$, which is divisible by $9$. Then assume $9 |(k^3+(k+1)^3+(k+2)^3)$ and try to prove that $9 |((k+1)^3+(k+2)^3+(k+3)^3)$ You will be fine if you can prove $9|((k+3)^3-k^3)$

Answer 3

Note than one of $n, n+1$ and $n+2$ will be a multiple of $3$ so it's cube will have a factor of $27=9*3$. The other two numbers will be congruent to $1$ and $2$ mod $3$ and so when cubed and added you will get a number congruent to $9$ mod $3$. In other words the entire sum is divisible by $9$.