Expansion of $C^j B^k$ for non-commutative $C$ and $B$ such that all $B$s are left of $C$s?

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Let $A$, $B$ and $C$ be some operators (details not relevant) and $\alpha$ a constant. The operators obey the following commutation relations:

$$ [A,B]=\alpha I $$ $$ [C,B]=2A$$ $$ [A,C]=0$$

I want to expand the following product

$$ C^j B^k $$

as a sum of terms where all $B$s are left of all $A$s and all $A$s are left of all $C$s, but I have no idea how to proceed... I tried expanding it by hand, but it gets complicated way too quickly and I can't spot a pattern. All help would be greatly appreciated!

By the way, $j,k \geq 0 $ and $j$ can be either less, equal, or greater than $k$.