Mathematics

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

$6$

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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Single Correct Medium Published on 09th 09, 2020
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