If α and β are the least positive integers such that for all n ∈ N,...

If for all $n \in N, n^3+\alpha n$ is divisible by 3 , then for $n=1$ also it is divisible by 3 , so the least positive integral value of $\alpha=2$ And, since for all $n \in N, n^3-\beta n$ is divisible by 6 , then for $n=1$ also it is divisible by 6 , so the least positive integral value of $\beta=1$. So, $\alpha+\beta=2+1=3$. Hence, option (b) is correct.