I have to prove that if $\displaystyle a,b,c\in\mathbb{Z}$ Show that $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1$ iff $\displaystyle gcd(a,lcm[b,c])=1$ I've tried proving the forward direction and im stuck. Here is what I have so far.

proof: $\displaystyle \rightarrow$: Assume $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1 $. Assume further that $\displaystyle gcd(a,[b,c])\neq 1 $. Let $\displaystyle d=(a,[b,c])$ which implies that $\displaystyle d|a$ and $d|[b,c] \implies d|b$ or $d|c $

Case 1: Suppose $\displaystyle d|b$ and $d\nmid c $. Since $\displaystyle d|a$ and $d|b$ and since $\displaystyle gcd(a,b)=1$ then d is a common multiple of $\displaystyle gcd(a,b)$ thus $\displaystyle d=1$ A contradiction.

Case 2: Suppose $\displaystyle d\nmid b$ and $d|c $. Same here as case 1 and d=1. A contradiction.

Case 3: Suppose $\displaystyle d|b$ and $d|c $ Since $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1$ and $d|a$,$\displaystyle d|b$, and $d|c $ then $\displaystyle d=1$ A contradiction.\

Would this be right?