# Divisibility Proof by Mathematical Induction: Sum of Three Consecutive Cubes

There are many ways to prove “The sum of three consecutive cubes is always divisible by nine”. This post explains how to determine the pattern of the sum of three consecutive cubes by listing the first few number series and to see how they relate to each other. Having some examples showing the sum of three consecutive cubes are always divisible by nine, and this will be proved by mathematical induction.

Before we actually proceed with the divisibility proof using mathematical induction, list some examples.

\begin{align} 1^3 + 2^3 + 3^3 &= 36 \\ &= 9 \times 4 \\ 2^3 + 3^3 + 4^3 &= 99 \\ &= 9 \times 11 \\ 3^3 + 4^3 + 5^3 &= 216 \\ &= 9 \times 24 \\ 4^3 + 5^3 + 6^3 &= 405 \\ &= 9 \times 45 \\ &\cdots \end{align}

$n^3 + (n+1)^3 + (n+2)^3 = 9M$, where $M$ is any positive integer

Thus it looks like the sum of three consecutive cubes numbers is always divisible by 9 by checking the above simple calculations.

## Proof by mathematical induction

Now let’s prove this statement using mathematical induction properly.

The first step is always to show the statement is true for $n = 1$, that is $1^3 + 2^3 + 3^3 = 36 = 9 \times 4$, which is divisible by $9$. This is true for $n=1$, the ignition step of the proof.

The second step of the proof is to assume the statement is true for $n = k$. That is $k^3 + (k+1)^3 + (k+2)^3 = 9P$, where $P$ is any positive integer.

The last step of the proof is to show the statement for $n=k+1$ is true based on the assumption for $n=k$ made at the second step above. That means the second step of the proof is to assume the statement is true for $n = k$. That is $(k+1)^3 + (k+2)^3 + (k+3)^3$ forms $9 \times$ any positive integer.

\begin{align} (k+1)^3 + (k+2)^3 + (k+3)^3 &= 9M – k^3 + (k+3)^3 &\color{green}{(k+1)^3 + (k+2)^3 = 9M – k^3 \text{ by the assumption}} \\ &= 9M – k^3 + k^3 + 9k^2 + 27k + 27 &\color{green}{\text{expand } (k+3)^3} \\ &= 9M + 9k^2 + 27k + 27 \\ &= 9(M+k^2 + 3k + 3) \end{align}

This tells us that the statement $(k+1)^3 + (k+2)^3 + (k+3)^3$ is idivsible by $9$. Thus the statement is true for $n=k+1$ based of the assumption $k^3 + (k+1)^3 + (k+2)^3$ is idivsible by $9$.

Therefore the statement “The sum of three consecutive cubes numbers is always divisible by 9” is true for any $n \ge 1$.

### What is the expression for the sum of three consecutive cubes?

$n^3 + (n+1)^3 + (n+2)^3$, for $n \ge 1$

### How do you find the sum of consecutive cubes?

The sum of the first $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.
$1^3 + 2^3 + 3^3 + \cdots + n^3 = (1+2+3+ \cdots +n)^2$

### Is the sum of three consecutive nth powers divisible by 15?

$4^n + 5^n + 6^n$ is idivsible by 15. You can prove this by mathematical induction. 