The image states:

"Cube of any integer is of the form 9k, 9k ± 1, k ∈ Z"

This statement refers to the property of cubes of integers when considered modulo 9. Let’s break it down:

Explanation:

For any integer $n$, when divided by 3, it will leave one of three possible remainders: 0, 1, or 2. Thus:

1. $n \equiv 0 \pmod{3}$
2. $n \equiv 1 \pmod{3}$
3. $n \equiv 2 \pmod{3}$

Now, consider the cubes of these values modulo 9:

1. If $n \equiv 0 \pmod{3}$:
   $n^3 \equiv 0^3 \equiv 0 \pmod{9}$.
   Form: $9k$.

2. If $n \equiv 1 \pmod{3}$:
   $n^3 \equiv 1^3 \equiv 1 \pmod{9}$.
   Form: $9k + 1$.

3. If $n \equiv 2 \pmod{3}$:
   $n^3 \equiv 2^3 \equiv 8 \pmod{9}$, and $8 \equiv -1 \pmod{9}$.
   Form: $9k - 1$.

Thus, the cube of any integer must be of the form $9k, 9k + 1, \text{ or } 9k - 1$ for some integer $k$.

Summary:

• $9k$: Cube of numbers divisible by 3.
• $9k + 1$: Cube of numbers leaving remainder 1 when divided by 3.
• $9k - 1$: Cube of numbers leaving remainder 2 when divided by 3.

This is why the cube of any integer is expressed as $9k, 9k \pm 1$, where $k \in \mathbb{Z}$ (set of all integers).

Questions for Further Understanding:

1. Why does the remainder analysis work modulo 9 for cubes?
2. How can this result be extended to other powers (e.g., $n^4$)?
3. Prove why $n^3$ modulo 9 is cyclic for $n = 0, 1, 2$.
4. Can this result be visualized using an example for integers like $n = 3, 4, 5$?
5. How is this property useful in number theory or modular arithmetic?

Tip:

Understanding modular arithmetic simplifies problems involving divisibility and remainders, especially when analyzing powers or exponents of integers.