Equivalence of outer lebesgue measure if covering of cubes is required rather than cuboids

Question

I want to prove that the outer lebesgue measure does not change when requiring to cover every set by cubes rather than cuboids. (i.e. See definition here: https://en.wikipedia.org/wiki/Lebesgue_measure) I know that the outer measure has to be at least as big when requiring cubes to cover the set, rather than cuboids since the set of all cubes is a subset of all cuboids. However, I have trouble showing that the measure is actually equal. I am kinda stuck at this point. I was however able to show that I can only take cubes with with diameter less than some $ \delta > 0 $ and get the same measure as for any cubes. Furthermore, I was able to show that I can take any elementary sets (a finite union of cubes) rather than cuboids to cover a set and also get the same outer measure value. Any help would be appreciated.