The difference between an integer and its cub | vedclass

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(B) Let the integer be $n$. The difference between the integer and its cube is given by $n^3 - n$.We can factorize this expression as: $n^3 - n = n(n^

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let the integer be $n$. The difference between the integer and its cube is given by $n^3 - n$.We can factorize this expression as: $n^3 - n = n(n^2 - 1) = n(n - 1)(n + 1)$.This is the product of three consecutive integers: $(n - 1)$,$n$,and $(n + 1)$.In any set of three consecutive integers,at least one must be a multiple of $2$ and exactly one must be a multiple of $3$.Since $2$ and $3$ are coprime,their product $2 	imes 3 = 6$ must divide the product of any three consecutive integers.Therefore,$n^3 - n$ is always divisible by $6$.

The difference between an integer and its cube is divisible by:

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