Let $a$ and $b$ denote integers.
To find $\frac{\mathrm{LCM}(a,b)}{a},$ we can use:
$$\frac{\mathrm{LCM}(a,b)}{a} = \frac{b}{\mathrm{HCF}(a,b)}$$ and compute $\mathrm{HCF}(a,b)$ with the Euclidean algorithm.
Question. Is there a similar formula that allows us to find $$\frac{\mathrm{LCM}(a,b,c)}{a}$$ using only arithmetical operations and highest common factors?
This is basically inclusion-exclusion: $$\text{LCM}(a,b,c)=\frac{abc\,\text{HCF}(a,b,c)}{\text{HCF}(a,b)\text{HCF}(a,c)\text{HCF}(b,c)}.$$
Sure, let's reinterpret the multiplicative monoid of $(\mathbb{Z}_{>0},\cdot)$ as $\newcommand{\NN}{\mathbb{N}}(\NN^\NN,+)$ via unique prime factorization. I.e. send a nonnegative integer to the sequence whose $n$th term is the exponent of the $n$th prime in its prime factorization. $\newcommand{\lcm}{\operatorname{lcm}}\newcommand{\gcd}{\operatorname{gcd}}$
Then lcm is just pointwise maximum in $\NN^\NN$, and likewise gcd is pointwise minimum. The fact that $ab=\gcd(a,b)\lcm(a,b)$ then translates to the fact that if $i,j\in\NN$, then $i+j=\max(i,j)+\min(i,j)$.
Now we need a version of this involving three natural numbers. $\max(i,j,k) + \min(i,j) + \min(j,k) + \min(i,k)$ will be the sum of the largest of all three, the middle of the three once, and the smallest of all three repeated twice. Hence $$\max(i,j,k) + \min(i,j) + \min(j,k) + \min(i,k) = i + j+ k+ \min(i,j,k).$$
Translating back into the multiplicative monoid, we have that the lcm of three integers satisfies $$\lcm(a,b,c) = \frac{abc \gcd(a,b,c)}{\gcd(a,b)\gcd(b,c)\gcd(a,c)}.$$
Another approach is to use$\newcommand{\lcm}{\operatorname{lcm}}$ $$\lcm(a,b,c) = \lcm(\lcm(a,b),c),$$ which can be generalized recursively to more elements. i.e.
$$\lcm(a_1,\lcm(a_2, \lcm(a_3,\cdots \lcm(a_{k-1},a_k) )))$$ This can be further broken down to a series of $\gcd(x,y)$'s which answers the original question. For example letting $d=\gcd(a,b)$, $$\lcm(a,b,c) = \lcm(\lcm(a,b),c) = \lcm(ab/d,c)=\frac{(ab/d)c}{\gcd(ab/d,c)}$$
Perhaps it may also be good to have a proof of the first statement, which is essentially same as jgon's answer: It suffices to consider the exponent of any prime $p$ for both sides and check that they are the same. Denote $v_p(x)$ the exponent of prime $p$ in the prime factorization of an integer $x$. For any chosen prime $p$, let $v_p(a)=u,v_p(b)=v,v_p(c)=w$. The $\lcm$ function ensures that the largest is chosen, so $$ \begin{align*} v_p(\lcm(a,b,c)) &= \max(v_p(a),v_p(b),v_p(c)) \\ &= \max(u,v,w) \\ v_p(\lcm(\lcm(a,b),c)) &= \max(v_p(\lcm(a,b)),v_p(c))\\ &= \max(\max(v_p(a),v_p(b)),w) \\ &= \max(\max(u,v),w)\\ &= \max(u,v,w) \end{align*} $$
