If y is an integer,then (y^{3}-y) is always a | vedclass

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(D) The given expression is $y^{3}-y$.We can factorize this expression as:$y^{3}-y = y(y^{2}-1)$Using the difference of squares formula $a^{2}-b^{2} =

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$6$

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(D) The given expression is $y^{3}-y$.We can factorize this expression as:$y^{3}-y = y(y^{2}-1)$Using the difference of squares formula $a^{2}-b^{2} = (a-b)(a+b)$,we get:$y^{3}-y = y(y-1)(y+1)$Rearranging the terms,we have:$(y-1) \cdot y \cdot (y+1)$This expression represents the product of three consecutive integers.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,$(y^{3}-y)$ is always a multiple of $6$.

If $y$ is an integer,then $(y^{3}-y)$ is always a multiple of:

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