Question : If $y$ is an integer, then $(y^{3}-y)$ is always multiple of _____.
Option 1: 5
Option 2: 7
Option 3: 9
Option 4: 6
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Correct Answer: 6
Solution :
$y^{3}-y=y(y^{2}-1)=y(y-1)(y+1)$
This product includes three consecutive integers: $y-1$, $y$, and $y+1$
In any set of three consecutive integers, at least one number is always a multiple of 3 and one is always even.
Hence, the correct answer is 6.
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