We want to find all integers $x$ between 1000 and 9999 that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is $10^{3}=1000$ and the largest such cube is $21^{3}=9261$. This means we only have to check 12 different cubes, which is quite doable, but we can reduce the search even further with a little number theory. Suppose we write our number as $x=1000 a+100 b+10 c+d$, where $a, b, c$, and $d$ are the decimal digits of $x$. Then we have $$(a+b+c+d)^{3} \equiv 1000 a+100 b+10 c+d \equiv a+b+c+d(\bmod 9)$$ If we let $k=a+b+c+d$, then $k$ must be a solution to the modular equation $k^{3} \equiv k(\bmod 9)$. A quick check of the values 0 through 8 shows that the only solutions are 0,1 , and 8 . Now, in our search, we only have to check values that are the cube of a number which is either 0,1 , or $8 \bmod 9$. $$\begin{aligned} & 10^{3}=1000, \text { but } 1+0+0+0 \neq 10 \\ & 17^{3}=4913, \text { and } 4+9+1+3=17 \\ & 18^{3}=5832, \text { and } 5+8+3+2=18 \\ & 19^{3}=6859, \text { but } 6+8+5+9 \neq 19 \end{aligned}$$ So the only solutions are 4913 and 5832 , which sum to 10745 .