Mathematics

Using Euclid's Division Lemma, for any positive integer $$n, n^3-n$$ is always divisible by 


ANSWER

$$6$$


SOLUTION
$$n^{3}-n=n(n^{2}-1)=n(n-1)(n+1)$$ is divisible by $$3$$ then  possible remainder is $$0, 1$$ and $$2$$

[$$\because$$ if $$P=ab+r$$, Then $$0\le r < a$$ by Euclid lemma]

$$\therefore$$ Let $$n=3r, 3r+1,3r+2$$ where $$r$$ is an integer

Case $$1$$: When $$n=3r$$

Then, $$n^{3}-n$$ is divisible by $$3$$  [$$\because n^{3}-n=n(n-1)(n+1)=3r(3r-1)(3r+1)$$, early shown it is divisible by $$3$$]

Case $$2$$: When $$n=3r+1$$

$$n-1=3r+1-1=3r$$

Then, $$n^{3}-n=(3r+1)(3r)(3r+2)$$ it is divisible by $$3$$

Case: when $$n=3r-1$$

$$m+1=3r-1+1=3r$$

Then, $$n^{3}-n=(3r-1)(3r-2)(3r)$$ it is divisible by $$3$$

Now out of three $$(n-1)^{n}$$ and $$(n+1)$$ are must be even so it is divisible by $$2$$

$$n^{3}-n$$ is divisible by $$2\times 3=6$$
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