Question:
For any positive integer $n$, prove that $n^{3}-n$ divisible by 6 .
Solution:
To Prove: For any positive integer $n, n^{3}-n$ is divisible by $6 .$
Proof: Let n be any positive integer.
$\Rightarrow n^{3}-n=(n-1)(n)(n+1)$
Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5
If n = 6q
Then, $(n-1) n(n+1)=(6 q-1) 6 q(6 q+1)$
If n = 6q + 1
Then, $(n-1) n(n+1)=(6 q+1)(6 q+2)(6 q+3)$
$\Rightarrow(n-1) n(n+1)=6(6 q+1)(3 q+1)(2 q+1)$
which is divisble by 6
Similarly we can prove others.
Hence it is proved that for any positive integer $n, n^{3}-n$ is divisible by 6 .
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