Assume that there exist $\overline{abcd}\in\mathbb{N}$ which $(a+b+c+d)^3=\overline{abcd}$ Thus $1000\leq (a+b+c+d)^3\leq9999\Rightarrow 10\leq a+b+c+d\leq21$ Since $\overline{abcd}\equiv a+b+c+d(mod~9)\Rightarrow (a+b+c+d)^3\equiv a+b+c+d(mod~9)$ We’ll get that $(a+b+c+d)((a+b+c+d)^2-1)\equiv0(mod~9)$ $\Rightarrow a+b+c+d\equiv 0,1,-1(mod~9)$ Hence, $a+b+c+d\in\{10,17,18,19\}$ We can check that $a+b+c+d\neq10,19$ and $a+b+c+d=17,18$ clearly works $\therefore$ the answers are $4913,5832 ~~~\square$