Question

Is it always true that difference of cube of any positive integer and the integer itself is always multiple of 3 ?​

Answer 1

Answer: Yes

Step-by-step explanation:

=$n^{3} -n = n(n^{2} -1)\\n(n+1)(n-1)$

Since n, n+1 and n-1 are consecutive numbers,

1 of them has to be multiple of 3.

Thus the statement is true...

Let's take integer 2.

$2^{3} = 8$

and, 8-2 = 6 which is multiple of 3.

Let's take integer 9.

$9^{3} = 729\\729-9 = 720$

And 720 is again multiple of 3.

Hope it helped.

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