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(B) For $n^3+\alpha n$ to be divisible by $3$ for all $n \in N$, we check $n=1$: $1^3+\alpha(1) = 1+\alpha$. For this to be divisible by $3$, the leas

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$3$

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(B) For $n^3+\alpha n$ to be divisible by $3$ for all $n \in N$, we check $n=1$: $1^3+\alpha(1) = 1+\alpha$. For this to be divisible by $3$, the least positive integer $\alpha$ is $2$ (since $1+2=3$).For $n^3-\beta n$ to be divisible by $6$ for all $n \in N$, we check $n=2$: $2^3-\beta(2) = 8-2\beta$. For this to be divisible by $6$, $8-2\beta = 6k$. If $\beta=1$, $8-2=6$, which is divisible by $6$. Thus, the least positive integer $\beta$ is $1$.Therefore, $\alpha+\beta = 2+1 = 3$.

If $\alpha$ and $\beta$ are the least positive integers such that for all $n \in N$, $n^3+\alpha n$ is divisible by $3$ and $n^3-\beta n$ is divisible by $6$, then $\alpha+\beta=$

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