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

$\mathrm{S}\left(\mathrm{n}\right)={\mathrm{n}}^3+\mathrm{\alpha } \mathrm{n}$ is divisible by

$\mathrm{S}\left(1\right)=1+\mathrm{\alpha }=3\mathrm{k} \Rightarrow \mathrm{\alpha }=3\mathrm{k}-1$

$\mathrm{S}\left(\mathrm{n}\right)={\mathrm{n}}^3-\mathrm{\beta } \mathrm{n}$ is divisible by 6

$\begin{array}{rcl}\mathrm{S}\left(1\right)& =& 1-\mathrm{\beta }=6\mathrm{m}\\ & \Rightarrow & \mathrm{\beta }=1-6\mathrm{m}\\ \mathrm{\alpha }+\mathrm{\beta }& =& 3\mathrm{k}-1+1-6\mathrm{m}\\ & =& 3(\mathrm{k}-2\mathrm{m})\end{array}$

is multiple of 3