$\operatorname{lcm}(2x, 3y-x, 3y+x)$ Problem
Find $\textrm{lcm} (2x, 3y-x, 3y+x)$, where $y > x > 0$ and $\gcd(x,y) = 1$.
Attempt
I noticed after some numerical calculation that the answer seems to depend on the parity of $x$ and $y$. However, I am basically at a loss of how to approach this. Any help would be appreciated.

You need to distinguish a few cases.
First note that if $3 \nmid x$, then $\gcd(3y - x, 3y + x)$ is either $1$ or $2$. Indeed, otherwise either a prime $p \neq 3$ or $4$ would divide both $6y$ and $2x$, but $x$ and $y$ are coprime by hypothesis. Furthermore, that $\gcd$ is $2$ if and only if both $x$ and $y$ are odd. Since no prime that divides $x$ can divide $3y-x$ or $3y+x$, again because $\gcd(x,y) = 1$, if $3 \nmid x$ we have
$$
\text{lcm}(2x, 3y-x, 3y+x) =
\begin{cases}
2x(3y-x)(3y+x) & \text{if } 2 \mid xy \\
\frac{1}{2}x(3y-x)(3y+x) & \text{otherwise}
\end{cases}
$$
Now suppose that $x = 3z$ for some $z \in \Bbb{Z}$. Since $\gcd(x,y) = 1$ no prime that divides $z$ can divide $y-z$ or $y+z$. Furthermore, just as above $\gcd(y-z,y+z)$ is $2$ if both $x$ and $y$ are odd, and it is $1$ otherwise. Thus if $3 \mid x$ we have
$$
\text{lcm}(2x, 3y-x, 3y+x) =
\begin{cases}
2x(y-z)(y+z) & \text{if } 2 \mid xy \\
\frac{1}{2}x(y-z)(y+z) & \text{otherwise}
\end{cases}
$$
