Prove that k ^ 3-k is divisible by 6 for any value of k.

We transform the expression k ^ 3 – k to the following form:

k ^ 3 – k = k * (k ^ 2 – 1) = k * (k – 1) * (k + 1) = (k – 1) * k * (k + 1).

The numbers k – 1, k and k + 1 are three integers in a row of integers one after the other

Let us show that one of them is necessarily divisible by 2 and one of them is necessarily divisible by 3.

If the number k – 1 is even, then it is divisible by 2, and if the number k – 1 is odd, then the number k is even and is divisible by 2.

If the number k – 1 is divisible by 3 with a remainder equal to zero, then it is divisible by 3.

If the number k – 1 is divisible by 3 with a remainder equal to 1, then the number k + 1 when divided by 3 will give a remainder equal to zero and will be divisible by 3.

If the number k – 1 is divisible by 3 with a remainder equal to 2, then the number k when divided by the number of three will give a remainder equal to zero and will be divisible by 3.

Thus, one of the numbers k – 1, k and k + 1 is divisible by 2 and one of these numbers is divisible by 3.

Therefore, either one of them is divisible by 6, or their product is divisible by 6.