smc

Note that $$\begin{array}{rl}{\Delta }^1(u_n)& =\left[(n+1{)}^3+(n+1)\right]-\left[n^3+n\right]\\ & =\left[n^3+3n^2+4n+2\right]-\left[n^3+n\right]\\ & =3n^2+3n+2,\\ {\Delta }^2(u_n)& =\left[3(n+1{)}^2+3(n+1)+2\right]-\left[3n^2+3n+2\right]\\ & =\left[3n^2+9n+8\right]-\left[3n^2+3n+2\right]\\ & =6n+6,\\ {\Delta }^3(u_n)& =\left[6(n+1)+6\right]-\left[6n+6\right]\\ & =\left[6n+12\right]-\left[6n+6\right]\\ & =6,\\ {\Delta }^4(u_n)& =6-6\\ & =0.\end{array}$$ Therefore, the answer is $\boxed{\text{if }k=4,\text{ but not if }k=3}.$ More generally, we have ${\Delta }^k(u_n)=0$ for all $n$ if $k\ge 4.$