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 $168$?

From $\sqrt[3]{abc}\approx 5.5 > 5\tfrac13=\frac{a+b+c}{3}$, we know that $a,b,c$ cannot all be positive (for in that case the AM-GM inequality states a "$\le $").
As $abc>0$, exactly two of $a,b,c$ must be negative.
Also, either all three numbers are even or exactly two are odd. In the first case, $\frac a2\frac b2\frac c2=21$, so all three integers $\frac a2,\frac b2,\frac c2$ are odd; but then $8=\frac a2+\frac b2+\frac c2$ would be odd as well. We conclude that exactly two of $a,b,c$ are odd, say $a,b$ are odd, $c$ is even. Wlog., $|a|\le |b|$. Then $c$ is an odd multiple of $8$, hence $a+b\equiv 8\pmod{16}$. The only ways to combine numbers from $\pm1,\pm3,\pm7,\pm21$ meeting this are $(a,b)=(\pm1,\pm7)$, or $(a,b)=(\pm3,\pm21)$; but the latter is not compatible with $ab\mid 168$. As two of $a,b,c$ must be negative, we conclude $$a=-1,\quad b=-7,\quad c=16-a-b=24$$
and from this
$$ x=\tfrac{b+c-a}2=9,\quad y=15,\quad z=-16.$$
