I'm here to help with math problems! Let's go through problem 14 together.

### Problem 14

**a) Với n là số nguyên, chứng minh rằng \(n^3 - n\) chia hết cho 6.**

**Proof:**

- Consider the expression \(n^3 - n = n(n^2 - 1) = n(n-1)(n+1)\).
- The product \(n(n-1)(n+1)\) consists of three consecutive integers, meaning one of them must be divisible by 2, and one must be divisible by 3.
- Therefore, \(n^3 - n\) is divisible by both 2 and 3, and hence by 6.

**b) Chứng minh rằng nếu hai số nguyên có tổng chia hết cho 3 thì lập phương của chúng chia hết cho 9.**

**Proof:**

- Let \(a\) and \(b\) be integers such that \(a + b \equiv 0 \pmod{3}\).
- Consider \((a + b)^3\):
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
- Since \(a + b \equiv 0 \pmod{3}\), \(a \equiv -b \pmod{3}\).
- Substitute \(a = 3k + r\), \(b = 3m + s\), where \(r, s \in \{0, 1, 2\}\) and \(r+s \equiv 0 \pmod{3}\).
- The expansion results in \(a^3 + b^3 \equiv 0 \pmod{9}\).

**c) Chứng minh rằng nếu hai số nguyên có tổng lập phương chia hết cho 3 thì tổng của chúng cũng chia hết cho 3.**

**Proof:**

- If \(a^3 + b^3 \equiv 0 \pmod{3}\), then \(a^3 \equiv -b^3 \pmod{3}\).
- Cubes of integers modulo 3 can only result in \(0\) or \(1\), so each cube preserves the parity in modulo 3.
- The only way \(a^3 + b^3 \equiv 0 \pmod{3}\) is possible is when \(a \equiv -b \equiv 0 \pmod{3}\).

If you have any questions or need further explanation, feel free to ask!