I am Equal to the Cube of my Digit Sum

We are looking for the numbers $N=abcd$ such that $abcd = (a + b + c + d)^{3}$ , with $10000 > abcd \ge 1000 \Rightarrow 21,54... = \sqrt[3]{10000} > a + b + c +d \ge 10 \Rightarrow \boxed{21 \ge a + b + c +d \ge 10}$ . Keep this in mind.

There are 3 possibilities:

$\boxed{1}$ .- $a + b + c + d = 3k - 1 \text{ ,with } k \in \mathbb{N} \Rightarrow (a + b + c + d)^{3} = (3k - 1)^{3} = 9m - 1$ , with $m \in \mathbb{N} \Rightarrow$ $(a + b+ c+ d)^{3} = abcd = 1000 \cdot a + 100 \cdot b + 10 \cdot c + d \equiv a + b + c + d \pmod {9}$ and $(a + b + c + d)^{3} \equiv -1 \pmod{9}$ $\Rightarrow a + b + c + d = 17 \Rightarrow 17^3 = \boxed{4913} \text{ is a number because } 21 \ge 4 + 9 + 1 + 3 = 17 \ge 10$ .

$\boxed{2}$ .- $a + b + c + d = 3k + 1 \text{ ,with } k \in \mathbb{N} \Rightarrow (a + b + c + d)^{3} = (3k + 1)^{3} = 9m + 1$ , with $m \in \mathbb{N}$ $\Rightarrow a + b + c + d = 10 \text{ or } 19 \Rightarrow 10^3 = 1000 \text{ is not a valid number because }$ $1 + 0 + 0 + 0 =1 < 10 \text{ and } 19^3 = 6859 \text{ is a not valid number because } 6 + 8 + 5 + 9 > 21$

$\boxed{3}$ .- $a + b + c + d = 3k \text{ ,with } k \in \mathbb{N} \Rightarrow (a + b + c + d)^{3} = (3k)^{3} = 9m$ , with $m \in \mathbb{N}$ $\Rightarrow a + b + c + d = 18 \Rightarrow 18^3 = \boxed{5832} \text{ is another number because }$ $21 \ge 5 + 8 + 3 + 2 = 18 \ge 10$ .

And there are not more numbers, therefore the solution is $5832 + 4913 = \boxed{10745}$