How to find the LCM of three numbers? We know that $LCM(a,b)= [\frac{ab}{\ GCD(a,b)}]$. What about $LCM (a,b,c)$? Can anyone help us because our instructors doesn't know the ways and she just lay the problem on us. Thanks.
 A: For 3 numbers, we have the relation
$$ LCM(a,b,c) = \frac{abc}{GCD(ab,bc,ca)} $$
The proof is as follows:
\begin{align}
a | n \text{ and } b | n \text{ and } c|n
& \iff abc | nab \text{ and } abc | nbc \text{ and } abc | nca\\
& \iff abc | GCD(nab,nbc,nca)\\
& \iff \frac{abc}{GCD(ab,bc,ca)} | n
\end{align} 
A: By considering prime factorizations, the statement $\text{lcm}(a,b,c)=\text{lcm}(\text{lcm}(a,b),c)$ reduces to:
$$
\max\{x,y,z\}=\max\{\max\{x,y\},z\}\qquad(\star)
$$
We prove this in two steps. First of all, it is clear that $\max\{x,y\}\leq \max\{x,y,z\}$ and $z\leq \max\{x,y,z\}$. Thus
$$
\max\{x,y,z\}\geq\max\{\max\{x,y\},z\}.
$$
On the other hand, we observe that either $\max\{x,y,z\}=\max\{x,y\}$ or $\max\{x,y,z\}=z$. Hence in either case,
$$
\max\{x,y,z\}\leq\max\{\max\{x,y\},z\}.
$$
Thus we've shown both inequalities, so equality $(\star)$ follows.

If you don't see why the statement $\max\{\max\{x,y\},z\}$ implies the $\text{lcm}$ statement, write out the prime factorizations of $a,b,c$. Fix any prime $p$, and compare the exponents appearing in $a,b,c$. Then the $\text{lcm}$ operation corresponds to taking the maximum of the exponents appearing in $a,b,c$ for the prime $p$.
A: For me, the easiest applied way (which can also easily be programmed in computer) to find it is using the prime factorization:
$$a_1,...,a_k\in\Bbb N\;,\;\;a_i=\prod_{m=1}^{n_i} p_{im}^{\beta_{im}}\;\;,\;p_{im}\;\;\text{primes,}\;,\;\;\beta_{im}\in\Bbb N$$
then
$$l.c.m.(a_1,...,a_n)=\prod_{i,m}p_{im}^{\max\limits_i(\beta_{i1},...,\beta_{ik})}$$
For example, if we have $\;15\;,\;\;72\;,\;\;32\;$ , then
$$\begin{cases}15=3\cdot5\\72=2^3\cdot3^2\\32=2^5\end{cases}\implies\;l.c.m.(15,72,32)=2^5\cdot3^2\cdot5$$
