Let $R$ be a UFD. Show that if $bc = ax$, then $b \in (a)$ or $c \in (a)$

Question

Let $R$ be a UFD. Let $b,c \notin U(R) = R^{\times}$, $b \neq 0 \neq c$ and let $a$ be an irreducible element such that $ax = bc.$ Prove that $b \in (a)$ or $c \in (a)$

My attempt:

Case 1: $x \notin U(R)$

$b,c,x$ have irreducible factorisations. Write them out:

$b = p_1 \dots p_n$

$c = q_1 \dots q_m$

$x = r_1 \dots r_l$

Then $$ax = a r_1 \dots r_l = p_1 \dots p_nq_1 \dots q_m = bc$$

and by unicity there exists $u \in U(R)$ such that $p_i = ua $ or $q_j = ua$. Hence, $b \in (a)$ or $c \in (a)$

Question 1: Is this correct?

Case 2: $x \in U(R)$

Here I'm stuck.

Question 2: Can someone hint me in the right direction?

Answer 1

Your first case is good. For the second, write $a = (x^{-1}b)c$. Then since $a$ is irreducible, either $x^{-1} b$ is a unit, or $c$ is. But then either $b = xc^{-1} a$ or $c = (x^{-1}b)^{-1}a$, and we're done.