Welcome 2016! Part 3

$\frac{x^{3}+y^{3}+z^{3}}{3}=xyz+21$

$x^{3}+y^{3}+z^{3}-3xyz=63$

We can factor the right side as follows: $(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)=(x+y+z)((x+y+z)^{2}-3(xy+yz+zx))$ We know that $x+y+z$ is a factor of 63. $x+y+z$ cannot be 1 or 3, because then the other factor is less than 63. $x+y+z$ cannot be 7, because then the second factor would have to be 9; however, the second factor is not divisible by 3. $x+y+z$ cannot be 9, because then the other factor is divisible by 3, which it cannot be. Now, let's see when $x+y+z=21$ . Then, $(x+y+z)^{2}-3(xy+yz+zx)=3$ , meaning that $xy+yz+zx=146$ .

By AM-GM, $xy+yz+zx\leq(\frac{x+y}{2})^{2}+(\frac{y+z}{2})^{2}+(\frac{z+x}{2})^{2}=(\frac{21-z}{2})^{2}+(\frac{21-x}{2})^{2}+(\frac{21-y}{2})^{2}\leq(\frac{21-7}{2})^{2}+(\frac{21-7}{2})^{2}+(\frac{21-7}{2})^{2}=147$

There is a chance that we can get $xy+yz+zx=146$ if we choose $(x,y,z)$ close to $(7,7,7)$ . Trying $(x,y,z)=(8,7,6)$ we see that it works. $\implies\boxed{x=8}$ .