GCD( LCM(a,b) , LCM(a,c) ) = LCM(a, GCD(b,c)) , How?

Question

I actually asked this question already once here, but I marked it answered by mistake and it was also titled wrong. I wanted to ask where I went wrong in the following lines and what should be the correct way to arrive at the conclusion. I suspect its line 3. $$ GCD\bigl(( LCM(a,b) , LCM(a,c)\bigr) \\\\=> a \times GCD\left( \frac{b}{GCD(a,b)} , \frac{c}{GCD(a,c)} \right) \\\\=> \frac{a \times GCD(b,c)}{GCD\bigl( GCD(a,b) , GCD(a,c)\bigr)} \\\\=> \frac{a \times GCD(b,c)}{GCD(a,b,c)} \\\\=> \frac{a \times GCD(b,c)}{GCD\bigl(a, GCD(b,c)\bigr)} \\\\=> LCM\bigl(a , GCD(b,c)\bigr) $$

Answer 1

HINT:

Use factorization. Let $a=p_1^{d_1}p_2^{d_2}\cdots p_i^{d_i}$, $b=p_1^{e_1}p_2^{e_2}\cdots p_i^{e_i}$, and $b=p_1^{f_1}p_2^{f_2}\cdots p_i^{f_i}$ (Some of the exponents may be $0$.)

Then, note that $\text{lcm}(a,b)=p_1^{\max\{d_1,e_1\}}p_2^{\max\{d_2,e_2\}}\cdots p_i^{\max\{d_i,e_i\}}$ and $\text{gcd}(a,b)=p_1^{\min\{d_1,e_1\}}p_2^{\min\{d_2,e_2\}}\cdots p_i^{\min\{d_i,e_i\}}$. Thus, to prove the equation, it is enough to prove $\min\{\max\{d_k,e_k\},\max\{d_k,f_k\}\}=\max\{d_k,\min\{e_k,f_k\}\}$ for each $k$.