since the equations are symmetric WLOG (1) eqn = product of (2)*3 we have also $a*b = \gcd(a,b)*lcm(a,b) \forall a,b $ so rewrite as $\gcd(a,b)*\frac{b*c}{\gcd(b,c)} =c*a\gcd(b,c)*\frac{a*b}{\gcd(a,b)}$ this implies $\gcd(a,b) = a*\gcd(b,c)$ $\gcd(a,b) =a*\gcd(b,c)\leq a$ $\gcd(b,c) = 1$ or we could said relative prime $\gcd(a,b) $indeed also exactly $ a$ implies that b exactly multiples of a

Another way to solve Let $\gcd(a,b) \cdot lcm(b,c) = \gcd(b,c)\cdot lcm(c,a) \gcd(c,a)\cdot lcm(a,b)$ or $(\gcd(a,b) \cdot lcm(b,c))^2 = \gcd(a,b) \cdot lcm(b,c) \gcd(b,c)\cdot lcm(c,a) \gcd(c,a)\cdot lcm(a,b)=a^2b^2c^2 \to \gcd(a,b) \cdot lcm(b,c)=abc\geq a \cdot lcm(b,c) \to \gcd(a,b) \geq a \to a|b$