elementary number theory - GCD property: $bmid ac$ implies $bmid (a,b)(b,c)$

The following is a very simple statement I want to prove:

If $a,b,c$ are non-zero integers, then $b\mid ac$ implies $b\mid (a,b)(b,c)$

Here $(a,b),[a,b]$ denote the greatest common divisor and the least common multiple between $a,b$, respectively.

The symbol $\mid$ means divisibility.

Attempt: $a,b,c\mid ac$ implies $[a,b]\mid ac, [b,c]\mid ac$.
Then $ac=t\frac{ab}{(a,b)}=u\frac{bc}{(b,c)}$ for some integers $t,u$.

Now, it follows that $(a,b)c=tb, (b,c)a=ub$.
Multiplying we get $(a,b)(b,c)ac=tub^2$.

Since $b\mid ac$, it follows the existence of an integer $q$ s.t. $ac=bq$, so that $(a,b)(b,c)bq=tub^2$.
Clearly $(a,b)(b,c)q=tub$.

Here I stuck.
Can I argue something more or should I follow a different strategy?

Thank you in advance for your help.

Update I would like to avoid Bezout identity, whenever possible, because I'm interested in applications to GCD domains, where a Bezout identity does not always hold. Some answers using Bezout were posted before this update.

Answer

$(b,a)(b,c)= ((b,a)b,(b,a)c) = (bb,ab,bc,ac) = b(b,a,c,ac/b)$