Is this a troll or something, this is trivial for Caucasus?
We use identity $\operatorname{lcm}(a,b) \cdot \gcd(a,b) =a b$ and we obtain that
$$\operatorname{lcm}(a,b)\cdot\operatorname{lcm}(b,c)\cdot\operatorname{lcm}(c,a)=\frac{ab}{\gcd(a,b)}\cdot \frac{bc}{\gcd(b,c)}\cdot \frac{ca}{\gcd(c,a)}=\frac{(abc)^2}{\gcd(a,b)\cdot\gcd(b,c)\cdot\gcd(c,a)}$$Since $(abc)^2$ and $\gcd(a,b)\cdot\gcd(b,c)\cdot\gcd(c,a)$ are perfect squares, then we conclude that $\operatorname{lcm}(a,b)\cdot\operatorname{lcm}(b,c)\cdot\operatorname{lcm}(c,a)$ is also a perfect square.