elementary number theory - Proving If and only if gcd(a,b) = gcd(b,c) = 1, then gcd(ab,c) = 1

I have tried to show this using the proposition that if gcd(a,b)=1, there exists two integers, x and y, such that $1 = ax + by$.

I first tried to prove the implication starting with gcd(a,b) = gcd(b,c) = 1.

$1 = ax + by$
and
$1 = bk + cm$

I then tried to manipulate these to the form
$1 = abl + cj$
(with l and j being integers)

I ran into trouble with multiplying by 1 in various places and replacing by either of the above identites because b is alone and both, so I kept on ending up with factors of b that I couldn't get rid of.

I also tried using the definition of gcd(a,b) = d, such that
$i) d > -1$ .
$ii)$ d divides both a and b.
$iii)$ any divisor of both a and b also divides d.

I got equally stuck using this method.

Answer

COUNTEREXAMPLE!

You cannot prove the title statement, it is false

Your title condition can be satisfied with $a=c,$ as long as the gcd with $b$ is one.

$$ a = 5,b=6, c = 5 $$

Then $$ \gcd(a,b) = 1$$
$$ \gcd(b,c) = 1$$

BUT
$$ \gcd(ab,c) = \gcd(30,5) = 5 $$