For any positive integer n, prove that n3 – n is divisible by 6.

1.	For any positive integer n, prove that n3 – n is divisible by 6.
Answer» Solution: n³ – n = n (n² – 1) = n (n – 1) (n + 1) Therefore, n³ – n is product of three consecutive positive integers, where n is any positive integer. Since one out of every two consecutive integers is divisible by 2. Therefore, The product n³ – n is divisible by 2. Since one out of every three consecutive integers is divisible by 3. Therefore, The product n³ – n is divisible by 3. Any number which is divisible by 2 and 3 is also divisible by 6. Hence, The product n³ – n is divisible by 6.

Answer»

Solution:
n³ – n = n (n² – 1) = n (n – 1) (n + 1)
Therefore, n³ – n is product of three consecutive positive integers, where n is any positive integer.
Since one out of every two consecutive integers is divisible by 2.

Therefore, The product n³ – n is divisible by 2.
Since one out of every three consecutive integers is divisible by 3.
Therefore, The product n³ – n is divisible by 3.
Any number which is divisible by 2 and 3 is also divisible by 6.
Hence, The product n³ – n is divisible by 6.

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