<s>[INST] Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$ 
Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$

Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?
 [/INST] We may even look out for integer solutions of $x^3+y^3+z^3=8$ alone. There is a huge literature on the sum of three cubes, and several data bases available. The equation has infinitely many solutions, for example
$$
\displaystyle (18t^3 + 2)^3 + (18t^4)^3 + (-18t^4 - 6t)^3 = 8
$$
for all integers $t$.
Among other solutions we find
$$
15^3+(-16)^3+9^3=8,
$$
which qualifies for $x+y+z=8$. Of course, we should solve the question as you did, but at least we already know a solution.
</s>