The sum of the cubes of three consecutive nat | vedclass

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(C) Let the three consecutive natural numbers be $(n-1), n, (n+1)$ where $n \geq 2$. The sum of the cubes of these numbers is: $(n-1)^3 + n^3 + (n+1)^

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the three consecutive natural numbers be $(n-1), n, (n+1)$ where $n \geq 2$. The sum of the cubes of these numbers is: $(n-1)^3 + n^3 + (n+1)^3 = (n^3 - 3n^2 + 3n - 1) + n^3 + (n^3 + 3n^2 + 3n + 1) = 3n^3 + 6n$. Factoring the expression,we get $3n(n^2 + 2)$. If $n$ is a multiple of $3$,then $3n$ is divisible by $9$. If $n$ is not a multiple of $3$,then $n^2 \equiv 1 \pmod{3}$,so $n^2 + 2 \equiv 1 + 2 \equiv 3 \equiv 0 \pmod{3}$. Thus,$n^2 + 2$ is divisible by $3$,making $3n(n^2 + 2)$ divisible by $3 	imes 3 = 9$. Therefore,the sum is always divisible by $9$.

The sum of the cubes of three consecutive natural numbers is divisible by

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