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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.

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{three}\:\mathrm{consecutive} \\ $$$$\mathrm{number}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\:\mathrm{9}. \\ $$

Answered by mr W last updated on 22/Apr/19

S=(n−1)^3 +n^3 +(n+1)^3 =n^3 −3n^2 +3n−1+n^3 +n^3 +3n^2 +3n+1 =3n^3 +6n =3n(n^2 +2) if n=3k: S=9k(9k^2 +2)=9×integer if n=3k±1: n^2 =9k^2 ±6k+1 ⇒S=3(3k±1)(9k^2 ±6k+3)=9(3k±1)(3k^2 ±2k+1) =9×integer

$${S}=\left({n}−\mathrm{1}\right)^{\mathrm{3}} +{n}^{\mathrm{3}} +\left({n}+\mathrm{1}\right)^{\mathrm{3}} \\ $$$$={n}^{\mathrm{3}} −\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{1}+{n}^{\mathrm{3}} +{n}^{\mathrm{3}} +\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1} \\ $$$$=\mathrm{3}{n}^{\mathrm{3}} +\mathrm{6}{n} \\ $$$$=\mathrm{3}{n}\left({n}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$${if}\:{n}=\mathrm{3}{k}: \\ $$$${S}=\mathrm{9}{k}\left(\mathrm{9}{k}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{9}×{integer} \\ $$$$ \\ $$$${if}\:{n}=\mathrm{3}{k}\pm\mathrm{1}: \\ $$$${n}^{\mathrm{2}} =\mathrm{9}{k}^{\mathrm{2}} \pm\mathrm{6}{k}+\mathrm{1} \\ $$$$\Rightarrow{S}=\mathrm{3}\left(\mathrm{3}{k}\pm\mathrm{1}\right)\left(\mathrm{9}{k}^{\mathrm{2}} \pm\mathrm{6}{k}+\mathrm{3}\right)=\mathrm{9}\left(\mathrm{3}{k}\pm\mathrm{1}\right)\left(\mathrm{3}{k}^{\mathrm{2}} \pm\mathrm{2}{k}+\mathrm{1}\right) \\ $$$$=\mathrm{9}×{integer} \\ $$

Commented by Tawa1 last updated on 22/Apr/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Tawa1 last updated on 22/Apr/19

Where do we see n = 3k±1 sir

$$\mathrm{Where}\:\mathrm{do}\:\mathrm{we}\:\mathrm{see}\:\:\mathrm{n}\:=\:\mathrm{3k}\pm\mathrm{1}\:\:\mathrm{sir} \\ $$

Commented by mr W last updated on 22/Apr/19

every number n is one of following three cases: n=3k n=3k+1 n=3k+2=3k−1 i.e. n=3k or n=3k±1

$${every}\:{number}\:{n}\:{is}\:{one}\:{of}\:{following} \\ $$$${three}\:{cases}: \\ $$$${n}=\mathrm{3}{k} \\ $$$${n}=\mathrm{3}{k}+\mathrm{1} \\ $$$${n}=\mathrm{3}{k}+\mathrm{2}=\mathrm{3}{k}−\mathrm{1} \\ $$$${i}.{e}.\:{n}=\mathrm{3}{k}\:{or}\:{n}=\mathrm{3}{k}\pm\mathrm{1} \\ $$

Commented by Tawa1 last updated on 22/Apr/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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