How do I find the integer solutions that satisfy $xyz = 288$ and $xy + xz + yz = 144$? 
Find all integers $x$, $y$, and $z$ such that $$xyz = 288$$ and $$xy + xz + yz = 144\,.$$

I did this using brute force, where $$288 = 12 \times 24 = 12 \times 6 \times 4$$ and found that these set of integers satisfy the equation. How do I solve this without using brute force?

We have $288 = 2^53^2.$  Let $x=2^a3^r,$ $y=2^b3^s,$ and $z=2^c3^t$.  Then $a+b+c = 5$ and $r+s+t=2$.  Since $r,s,$ and $t$ are non-negative integers, one of them must be $0$, say $t=0$.  From the equation
$$xy+xz+yz = 144,$$
we see that if a prime number divides any of the variables, it must divide at least one of the others.  This forces $r =s=1$ and we must have
$$x=2^a3, y=2^b3, z=2^c.$$
Similarly, if one of the variables is divisible by $8$, then the product of the other two variables is also divisible by $8$, but then $a+b+c\geq 6$, which is too big.  This forces $1\leq a,b,c \leq 2.$
So either $c=1$ or $c=2.$  If $c=1$, then $a=b=2$ and $x=y=12$ and $z=2$, which doesn't satisfy the second equation.
If $c=2$, then $a=2, b=1$ (or vv.) and we have $x=12$, $y=6$, $z=4$ which is the only solution.
