If $ a \mid bc $ then $\frac{a}{\gcd(a,b)} \mid c$? Prove or reject this statement:

If $ a \mid bc $ then $\displaystyle \frac{a}{\gcd(a,b)} \mid c$


Hint: If $a\mid bc$ and $(a,b)=1$ then $a\mid c$. Use this to show:
$$
\displaystyle  a \mid bc \implies \frac{a}{\gcd(a,b)} \mid \frac{b}{\gcd(a,b)}\times c \implies \frac{a}{\gcd(a,b)}\mid c
$$
Because $\gcd(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)})=1$.
