Determine the values of x, y, and z using one equation with three variables Solve for x, y, and z using one equation with three variables:
$xyz = x! + y! + z!$ 
where factorials are defined in the usual way, namely $n! = n(n-1)...2*1$
I'm not sure where to start with this question, any help would be great! Thanks!

$xyz=x!+y!+z!$
$AM \ge GM$.
$\therefore$ $x^3+y^3+z^3\ge3x!+3y!+3z!$
$\Rightarrow$ $x^3+y^3+z^3-(3x!+3y!+3z!)\ge0$ -------- (1)
But we know that $k^3<k!$ (see above comment by JMoravitz) for $k>5$.  
$\max(n^3-3n!)=9, n=3$. --------- (2)  
That means that if we try to take value of $(x,y,z)$ greater than $4$ we won't be able to keep the above expression (expression (1), above) greater than $0$. As for $n=5$, expression(expression (2)) gives value smaller than $-18$ (which is the minimum value we can cancel out (by putting $n=3$) i.e. $18=2*9$).
Note: Expression (2) will keep on further decreasing for higher values of $n$.
Hence, it proves that $n\le4$.  
Now find, value of expression (2) for $n=1,2,3,4$. ($0$ ins't possible because LHS will turn $0$).
Values of expression (2),
For $n=1$, Value $=-2$
$n=2$, Value $=2$
$n=3$, Value $=9$
$n=4$, Value $=-8$.  
All possible triplets are $(1,2,2);(1,3,3);(1,2,3);(4,3,2);(4,3,3);(3,3,3);(2,2,2);(2,2,3);(3,3,2)$  
Indeed a great achievement from $64$ possible triplets to $10$ (could be done faster if you have intuition and practised these steps.)  
Bonus: $(x,y,z)$ aren't equal because for that $(x,y,z)$ should have $3$ as a factor and should be even. Reducing triplets to $8$.  
This isn't answer you have to solve. :P
