Question
For any positive integer n, prove that $n^3 - n$ is divisible by $6.$
✓
Answer
Let $n = 6q or 6q + 1, 6q + 2, 6q + 3 ... 6q + 5$
Hence we can similarly, prove that $n^2 - n$ is divisible by $6$ for any positive integer n.
Hence proved.
If $n = 6q$, then
Then $n^3 - n = (6q)^3 - 6q = 216q^3 - 6q$
$= 6 (36q^3 - q)$
Which is divisible by 6
If $n = 6q + 1$, then
$n^3 - n = (6q + 1)^3 - (6q + 1)$
$= 216q^3 + 108q^2 + 18q + 1 - 6q - 1$
$= 216q^3 + 108q^2 + 12q$
$= 6(36q^3 + 18q^2 + 2q)$
Which is also divisible by 6
If $n = 6q + 2,$ then
$n^3 - n = (6q + 2)^3 - (6q + 2)$
$= 216q^3 + 216q^2 + 72q + 8 - 6q - 2$
$= 216q^3 + 216q^2 + 66q + 6$
$= 6(36q^3 + 36q^2 + 11q + 1)$
Which is divisible by $6$Hence we can similarly, prove that $n^2 - n$ is divisible by $6$ for any positive integer n.
Hence proved.
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