How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3. 
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?

Since you have proven that this formula is available for $n=1$ we suppose that formula is available for all natural numbers n=k
$$k(k^2+8)$$ then according to axiomme of mathematical induction we need to prove that formula is valid for $n=k+1$ or 
$$(k +1)((k+1)^2+8)$$ is multiple of 3
now we have 
$$(k+1)((k+1)^2+8)=(k+1)(k^2+2k+1+8)=k^3+3k^2+11k+9=(k^3+8k)+(3k^2+3k+9)=k(k^2+8)+3(k^2+k+3)$$
first part of expression is factor of 3 by assumption an second part evidently is factor of 3
