Proving 9 divides a cubic by Induction

Question

I have just started to cover induction mathematics in my Discrete Mathematics class and I'm a little confused as to where to go with this problem. Am I on the right track?

Prove that 9 divides (n^3 + (n+1)^3 + (n+2)^3) when using some integer k >= 0

Basis case: n=0, n=1

(0^3 + (0+1)^3 + (0+2)^3)/9 = (0 + 1 + 8)/9

= 9/9 = 1 TRUE

(1^3 + (1+1)^3 + (1+2)^3)/9

= (1 + 8 + 27)/9

= 36/9 = 4 TRUE

Induction hypothesis: For fixed number k >= 0, S(k): is true where:

(k^3 + (k+1)^3 + (k+2)^3)/9

Prove that S(k+1) is true for all numbers k >= 0:

S(k+1): ((k+1)^3 + ((k+1)+2)^3 + ((k+1)+3)^3)/9

((k+1)(1+2+3)^3)/9

((k+1) * 9^3)/9

Im not sure where to go from here.

Answer 1

Non-inductive proof: $n^3+(n+1)^3+(n+2)^3=3\underbrace{(n-1)n(n+1)}_{3\text{ consecutive integers}}+\underbrace{9(n^2+2n+1)}_{\text{divisible by }9}$

Three consecutive integers must have one of them divisible by $3$.

You wrote $\frac{(k+1)^3 + ((k+1)+2)^3 + ((k+1)+3)^3}{9}=\frac{(k+1)(1+2+3)^3}{9}$, which is incorrect. Also then you wrote $1+2+3=9$.

$$\begin{align}(k+1)^3+(k+2)^3+(k+3)^3&=k^3+(k+1)^3+(k+2)^3+(k+3)^3-k^3\\&=\underbrace{k^3+(k+1)^3+(k+2)^3}_{\text{inductive hypothesis}}+\underbrace{9 (k^2+3 k+3)}_{\text{ divisible by }9}\end{align}$$

Answer 2

Hint: Recall that $(a+b)^3=a^3+3a^2b+3ab^2+b^3$.

Expand $(k+1)^3 + ((k+1)+1)^3 + ((k+1)+2)^3$ to get $$(k^3+(k+1)^3+(k+2)^3)+\text{something}$$

Study that something.