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

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

Facebook Tweet Pin