<s>[INST] Find the sum of the digits of the four-digit number 
It is known that $\overline{abcd}-\overline{dcba}=x$ and $a^3+b^3+c^3+d^3=x$. Find $a+b+c+d$.

From the given we get $999a+90b-90c-999d = a^3+b^3+c^3+d^3$. We could just test out the possibilities, but is there another way to find the sum of the digits?
 [/INST] Partial Answer
As $x=999a+90b-90c-999d=9(111a+10b-10c-111d)$ then $x\equiv0\pmod{9}$. Looking at cubes modulo 9 we see:
$0^3\equiv0\pmod{9}$, $1^3\equiv1\pmod{9}$, $2^3\equiv8\pmod{9}$, $3^3\equiv0\pmod{9}$, $4^3\equiv1\pmod{9}$, $5^3\equiv8\pmod{9}$, $6^3\equiv0\pmod{9}$, $7^3\equiv1\pmod{9}$, $8^3\equiv8\pmod{9}$, $9^3\equiv0\pmod{9}$.
For the four cubes to be a multiple of $9$ we require:


*

*4 zeros $\big((a,b,c,d)\equiv(0,0,0,0)\pmod{3}\big)$, or

*2 zeros, a one and an eight $\big((a,b,c,d)\equiv(0,0,1,2)\pmod{3}\big)$, or

*2 ones, 2 eights $\big((a,b,c,d)\equiv(1,1,2,2)\pmod{3}\big)$


We have $999(a-d) = a^3+b^3+c^3+d^3+90(c-b)$. Next consider the maximum value this can take. This occurs when $(a,b,c,d)=(9,0,9,9)$. The right hand side in this case is equal $2997$ (with equality only if $a=d$). Thus $0\le a-d\le2$.
I'll keep adding to this...
Cheating
Via computer the answer is $a=4$, $b=6$, $c=8$, $d=3$. I'll keep thinking about how to get here without cheating. :) 
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