prove ( n∈ N ) 3(n+1) ∣ n^( 3) + (n+1)^( 3) + (n+2 )^( 3)


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Question Number 165329 by mnjuly1970 last updated on 30/Jan/22

$p r o v e (n \in N)$ $3 (n + 1) ∣ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}$
	Answered by Rasheed.Sindhi last updated on 30/Jan/22
	$3(n+1) ∣ n^( 3) + (n+1)^( 3) +(n+2 )^( 3) _(−) ∵ n^( 3) + (n+1)^( 3) +(n+2 )^( 3) =(n+1−1)^3 + (n+1)^( 3) +(n+1+1 )^( 3) =(n+1)^3 −1−3(n+1)(n+1−1) +(n+1)^3 +(n+1)^3 +1+3(n+1)(n+1+1) =3(n+1)^3 −3n(n+1)+3(n+1)(n+2) =3(n+1){(n+1)^2 −3n+n+2} ∴ 3(n+1) ∣ n^( 3) + (n+1)^( 3) +(n+2 )^( 3)$
	$\underline{3 (n + 1) ∣ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}}$ $∵ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}$ $= {(n + 1 - 1)}^{3} + {(n + 1)}^{3} + {(n + 1 + 1)}^{3}$ $= {(n + 1)}^{3} - 1 - 3 (n + 1) (n + 1 - 1)$ $+ {(n + 1)}^{3}$ $+ {(n + 1)}^{3} + 1 + 3 (n + 1) (n + 1 + 1)$ $= 3 {(n + 1)}^{3} - 3 n (n + 1) + 3 (n + 1) (n + 2)$ $= 3 (n + 1) {{(n + 1)}^{2} - 3 n + n + 2}$ $∴ 3 (n + 1) ∣ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}$
	Commented by mnjuly1970 last updated on 30/Jan/22

	$v e r y n i c e s o l u t i o n s i r . . .$
	Answered by Rasheed.Sindhi last updated on 31/Jan/22
	$AnOther method 3(n+1) ∣ n^3 + (n+1)^3 + (n+2 )^3 _(−) ∵ n^3 + (n+1)^3 + (n+2 )^3 =a^3 +b^3 +c^3 −3abc+3abc ; { ((a=n)),((b=n+1)),((c=n+2)) :} =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)+3abc =(n+n+1+n+2)( ∼ )+3n(n+1)(n+2) =3(n+1)( ∼ )+3n(n+1)(n+2) =3(n+1){( ∼ )+n(n+2)} ∴ 3(n+1) ∣ n^3 + (n+1)^3 +(n+2 )^3$
	$AnOther method$ $\underline{3 (n + 1) ∣ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}}$ $∵ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}$ $= a^{3} + b^{3} + c^{3} - 3 a b c + 3 a b c; {\begin{cases} a = n \\ b = n + 1 \\ c = n + 2 \end{cases}$ $= (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c$ $= (n + n + 1 + n + 2) (\sim) + 3 n (n + 1) (n + 2)$ $= 3 (n + 1) (\sim) + 3 n (n + 1) (n + 2)$ $= 3 (n + 1) {(\sim) + n (n + 2)}$ $∴ 3 (n + 1) ∣ n^{3} + {(n + 1)}^{3} + {(n + 2)}^{3}$
	Commented by mnjuly1970 last updated on 31/Jan/22

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