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:
Let the highest power of prime $p$ that divides $a,b,c$ be $A,B,C$ respectively
As $a|bc, A\le B+C\iff A-B\le C\  \  \  \ (1) $
Now, the highest power$(D)$  of prime $p$ that divides $\displaystyle \frac a{(a,b)}=A-$min$(A,B)$
If $A\ge B,D=A-B\le C$ by $(1)$
Else $D=A-A=0$ which is always $\le C$
