Check: For the integers $a,b,c$ show that $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$

For the integers $a,b,c$ show that $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$

Proof:

Let $u$ and $v$ be integers. Then $\gcd(a,b)=au+bv$. Then $c\cdot \gcd(a,b)=c\cdot(au+bv)=acu+bcv$ Let $x$ and $y$ be integers. Then $\gcd(a,\gcd(a,b)\cdot c)=ax+(acu+bcv)y=ax+acuy+bcvy=a(x+cuy)+(bc)(vy)$.

Let $d$ and $e$ be integers. Then $\gcd(a,bc)=ad+bce$.

Since $d$ and $e$ are integers let $d=x+cuy$ and $e=vy$. Hence $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$

${\begin{eqnarray}{\bf Hint}\ \ \ &&(a,\quad\ (\ a,\ \ \ \ b)\ c) &=&\ (a,\ \ \ ac, \ \ \ \ \ bc) &=& ((a,\ \ \ ac),\ \ \ \ \ bc) &=&\ (a,\ \ \ bc)\\[.3em] {\rm or}\ \ \ &&\ a\Bbb Z + (a\Bbb Z\!+\!b\Bbb Z)c &=&\ \ a\Bbb Z\!+\!ac\Bbb Z\!+\!bc\Bbb Z &=&\ (a\Bbb Z\!+\!ac\Bbb Z)\!+\!bc\Bbb Z&=&\ \ a\Bbb Z\!+\!bc\Bbb Z\end{eqnarray}}\,$

The first proof uses GCD laws (distributive, commutative, associative); the second uses Bezout.

Remark $\ $ The gcd arithmetic in the first proof can be made more intuitive by using a notation that exploits its analogy with ordinary integer arithmetic, viz. denote the gcd $\rm\:(a,b)\:$ by $\rm\ a \dot+ b,\ $ so

Lemma $\rm \ \ (ac,b) = ((a,b)c,b)\,\ [=\, (c,b)\ \ {\bf if}\ \ (a,b)=1]$

$\begin{align}\rm {\bf Proof}\ \ \ ((a,\,b)c,\ b) &\rm =\, (ac,\,bc,\,b) = (ac,\ b\color{#c00}{(c,\ 1)}) = (ac,b)\\[.3em] \rm (a\dot+b)c\dot+b\ \:\! &\rm =\,\ ac\dot+bc\dot+b\, = \ \, ac\dot+b\color{#c00}{(c\dot+1)}\ =\ ac\dot+b \end{align}$

Because the arithmetic of GCDs shares many of the same basic laws of the arithmetic of integers, the proof becomes more intuitive using this notation that highlights this common arithmetical structure. The proof employs only said basic GCD laws, namely the associative law, and the commutative law, combined with the fundamental distributive law $\rm\, \ (ac,bc)\ =\ (a,b)\:c,\ $ and the GCD-specific law $\rm\: \color{#C00}{(1,c) = 1}.\ $ For a less trivial example see my similar proof of the Freshman's Dream $\rm\ (A,B)^n\, =\ (A^n,B^n)\ $ for GCDs and cancellable ideals.

The motivation behind this powerful abstract axiomatic approach becomes much clearer should you go on to study ideal theory and/or divisor theory (and you should, for there lies much beauty).

For further discussion and generalizations see this post and see my various posts on GCDs.

Since a proof has already been sketched by Bill, let me give you an alternative solution/approach to the problem that you mind find intuitive.

First off, look at the formula that you're asked to establish: $$ \gcd(a,bc) = \gcd(a,\gcd(a,b)\cdot c). $$ Can you translate this equation into a simple English sentence? Really think about what it means given the meaning of GCD. Here's my attempt, which is actually a slightly stronger statement:

A divisor of $a$ also divides $bc$ if and only if part of it divides $b$ and the rest of it divides $c$.

Thinking about it in these terms suggests an elementary proof. (I'll let you fill in any small gaps below.) Let $d = \gcd(a,b)$. Then $a = da'$ and $b = db'$ for relatively prime integers $a',b'$. Thus $\gcd(a,bc) = \gcd(da',db'c) = d\gcd(a',b'c)$. Now to say that $a'$ and $b'$ are relatively prime means that they share no common prime factors. So a prime $p$ divides both $a'$ and $b'c$ if and only if it divides $a'$ and $c$. As a consequence $\gcd(a',b'c) = \gcd(a',c)$, and therefore $D = \gcd(a,bc) = d\gcd(a',b'c) = d\gcd(a',c) = \gcd(da',dc) = \gcd(a,dc)$ as claimed.