Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade. What do you think about my first induction proof? Please mark/grade.

Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows.
        $$\forall n \in \mathbb{N}, P(n)
  \quad \text{where} \quad
  P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$
        We prove the theorem by induction on $n$.
Basis
Below, we show that we have $P(n)$ for $n = 0$.
            $$0^3 + 1^3 + 2^3 = 0 + 1  +  8 =  9 = 9 \cdot  1$$
Inductive step
Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$.
Let $k \in \mathbb{N}$. We assume that $P(k)$ holds.
            In the following, we use this assumption to show that $P(k + 1)$ holds.
By the assumption,
            there is a $i \in \mathbb{N}$ such that
            $i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$.
            We use this fact in the following equivalent transformation.
            The transformation turns the sum of cubes in the first line,
            for which we need to show that it is a multiple of 9,
            into a product of 9 and another natural number.
$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\
   = (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\
   = k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\
   = 9i + 9k^2 + 27k + 27 \\
   = 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\
   = 9 \cdot (i + k^2 + 3k + 3)$
We see that the above product has precisely two factors: 9 and another natural number.
            Thus the product is a multiple of 9.
            This completes the induction.

Formulation, base case, inductive hypothesis, inductive step, it all looks good. :) 
One might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \in \Bbb N$.
