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 $168$?

Hint: 
Taking from where you left off: $ab \mid 168 \implies ab = \pm 1, \pm 2, \pm 4, \pm 6, \pm 7, \pm 8, \pm 12, \pm 14, \pm 21, \pm 24, \pm 28, \pm 42, \pm 56, \pm 84, \pm 168$. Even though it looks cumbersome, it is easy to solve. For example, $ab = 6 \implies c = \dfrac{168}{6} = 28 \implies a+b = 16-c = 16 - 28 = -12$. This case yields no solution. But other cases yield some solutions as you go through them you might find the work is not that tedious and may be fun indeed.
