Prove or disprove statements about the greatest common divisor Help with prove or disproving either of these statements would be really appreciated, one or the other is fine, I just need a start or a solution to one and I'm sure I could probably figure the other out.

(a) For all $a, b, c \in \mathbb{Z}$, $\gcd(ac, bc) = c \gcd(a, b)$
(b) For all $a, b, c \in \mathbb{Z}$, $\gcd(ab, c) = \gcd(a, b) \gcd(b, c)$

 A: Hint for (a): Note that $c|ac$ and $c|bc$.
Hint for (b): Look at special cases, $a=b$, $a=c$ etc
A: I find it helpful to look at the prime factor decomposition of the numbers $a,b$ and $c$. For two numbers $x,y$ the $\gcd(x,y)$ is just the maximal set of common prime factors. 
So for (a) you it is clear that the maximal sets of prime factors of $ac$ and $bc$ contain  both the prime factors of $c$. So if you take the prime factors of $c$ out, both sets remain maximal.
For (b) I would suggest to play around with some examples.
A: If the highest power of prime $p$ in $a,b,c$ are $r_a,r_b,r_c$ respectively, 
the highest power of $p$ in gcd$(a,b)=$min $(r_a,r_b)$
the highest power of $p$ in $c\cdot$ gcd$(a,b)=r_c+$min $(r_a,r_b)$
the highest power of $p$ in  gcd$(a\cdot c,b \cdot c)=$min $(r_a+r_c,r_b+r_c)=r_c$+min $(r_a,r_b)$
the highest power of $p$ in gcd$(a\cdot b,c)=$min $(r_a+r_b,r_c)$
the highest power of $p$ in gcd$(a,b)\cdot $gcd $(b,c)=$min $(r_a,r_b)$ + min$(r_b+r_c)$
Can you prove min $(r_a+r_b,r_c)=$min $(r_a,r_b)$ + min$(r_b+r_c)?$
A: Concerning (b), perhaps it's worth mentioning that $\gcd$ and $\operatorname{lcm}$ distribute each with respect to the other, so you have
$$
\begin{align}
&\gcd(\operatorname{lcm}(a, b), c) = \operatorname{lcm}(\gcd(a, c), \gcd(b, c)),
\\&
\operatorname{lcm}(\gcd(a, b), c) = \gcd(\operatorname{lcm}(a, c), \operatorname{lcm}(b, c)).
\end{align}
$$
The first of these equalities is the correct form of (b).
