**For any positive integer n, prove that n ^{3} – n is divisible by 6.**

**Solution: **when (n^{3} – n) is divided by 6, then (n^{3} – n) = n (n -1) (n +1) is also divided by 6.

(n^{3} – n) = n (n -1) (n +1)

** **If a number is divisible by both 2 and 3, then the number is divisible by 6. When a number (n) is divided by 3, to get q as quotient and r as remainder.

By Euclid’s division lemma, we get

n = 3q + r, where 0 ≤ r < 3.

All possible remainder = 0, 1, 2

**Case I: **When r = 0

n = 3q + r = 3q + 0 = 3q is divisible by 3.

**Case II: **When r = 1

n = 3q + 1, then n-1 = 3q + 1 – 1 = 3q is divisible by 3.

**Case III: **When r = 2

n = 3q + 2, then n+1 = 3q + 2 + 1 = 3q + 3 = 3(q+1) is divisible by 3.

We can say that n, (n-1) and (n+1) is divisible by 3. Then n (n-1) (n+1) is always divisible by 3.

Similarly, when a number is divided by 2 to get q as quotient and r as **remainder**.

n = 2q + r, where 0 ≤ r < 2.

All possible remainder = 0, 1

**Case I: **when r = 0

** **n = 2q + 0 = 2q is divisible by 2.

**Case II: **When r = 1

** **n = 2q + 1

n – 1 = 2q + 1 – 1 = 2q is divisible by 2.

And n + 1 = 2q + 1 + 1 = 2q + 2 = 2(q+1) is divisible by 2.

We can say that n, (n-1) and (n+1) is divisible by 2 then n (n-1) (n+1) is also divisible by 2.

Since, n (n-1) (n+1) is divisible by both 2 and 3.

Hence, n (n-1) (n+1) = n^{3} – n is divisible by 6.

**SOME OTHER IMPORTANT QUESTIONS**

**Use Euclid’s division lemma to show that the cube of any positive integer is of form 9m, 9m+1, and 9m+8.****State Fundamental theorem of arithmetic.**

**Related Questions **

**Prove that for any positive integer n,n3−n is divisible by 6.**

**How can we prove that n³-n is divisible by 6 if n is any positive integer?**

**If n is a positive integer, is n³-n divisible by 6?**