Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$ [duplicate]

Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$. My attempt is let $\gcd(ab,c)=d$. Since $d \mid ab$ and $d \mid c$ , $d \mid (abt+cs)$ for some integers $s$ and $t$. Then by definition of divisibility, we have $$abt+cs=dk$$ for some integers $k$. Then I got stuck here.

Hint 1: First approach. Assume by contradiction that $d \neq 1$. Pick some prime $p$ which divides $d$. Then $p\mid ab$ and $p\mid c$. Do you see the contradiction?

Hint 2: Second approach. You know that

$$ax+cy=1$$ $$bz+ct=1$$

For some positive integers.The second equation yields

$$abz+act=a$$

Plugging in the first equation you get

$$(abz+act)x+cy=1$$

But this implies that $\gcd(ab,c)=1$ (Why?)...

Using the fundamental theorem of arithmetic, if $a$ and $c$ are coprime then there is no common prime in their prime factorisations. The same goes for $b$ and $c$, and hence the same goes for $ab$ and $c$.