Show that the sum of the cube of three consecutive number gives a multiple of 9.


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Question Number 58410 by Tawa1 last updated on 22/Apr/19

$Show that the sum of the cube of three consecutive$ $number gives a multiple of 9 .$
	Answered by mr W last updated on 22/Apr/19

	$S = {(n - 1)}^{3} + n^{3} + {(n + 1)}^{3}$ $= n^{3} - 3 n^{2} + 3 n - 1 + n^{3} + n^{3} + 3 n^{2} + 3 n + 1$ $= 3 n^{3} + 6 n$ $= 3 n (n^{2} + 2)$ $i f n = 3 k :$ $S = 9 k (9 k^{2} + 2) = 9 \times i n t e g e r$ $i f n = 3 k \pm 1 :$ $n^{2} = 9 k^{2} \pm 6 k + 1$ $\Rightarrow S = 3 (3 k \pm 1) (9 k^{2} \pm 6 k + 3) = 9 (3 k \pm 1) (3 k^{2} \pm 2 k + 1)$ $= 9 \times i n t e g e r$
	Commented by Tawa1 last updated on 22/Apr/19

	$God bless you sir$
	Commented by Tawa1 last updated on 22/Apr/19

	$Where do we see n = 3 k \pm 1 sir$
	Commented by mr W last updated on 22/Apr/19

	$e v e r y n u m b e r n i s o n e o f f o l l o w i n g$ $t h r e e c a s e s :$ $n = 3 k$ $n = 3 k + 1$ $n = 3 k + 2 = 3 k - 1$ $i . e . n = 3 k o r n = 3 k \pm 1$
	Commented by Tawa1 last updated on 22/Apr/19

	$God bless you sir$
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