Order of operations and allowance of shortcuts

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Order of operations and the BODMAS are fundamental to all tenets of maths, but, in the interest of time-saving, can shortcuts be allowed, especially one when already knows the end result?

Supposed I would like to re-write the expression $a[b(c-d)]$ as $ab(c-d)$.

I could of course immediately do so i.e. $a[b(c-d)]$ = $ab(c-d)$ by multiplying $a$ in directly.

However, this might not strictly follow the BODMAS rule because I did not deal with what was in the inner brackets first. If I follow the BODMAS rule which means I must tackle what's in inner the brackets first, I would write:

$a[b(c-d)]=a(bc-bd)=abc-abd=ab(c-d)$

This however, seems unnecessarily trivial and time-wasting at college-level mathematics especially when I know what the end-result will already be.

Would like to seek advice on this - must the long way always be taken?

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IMHO, order of operations is indeed pretty fundamental — in the sense of being the spelling rules, so to speak, to ensure that those who write and those who read mathematical expressions can understand, not misunderstand, each other.

"BODMAS", and its North American counterpart "PEMDAS", are rather detrimental, not fundamental, to math. (I can't believe I've just typed those two with my own fingers…) It's a pedagogical disaster, if you ask me. And your example is quite a nice illustration to this point! Another example of the harm they do is the widely popular misconception among schoolchildren and college students that "addition should always be performed before subtraction because A is before S in you know where". And equipped with this "knowledge" they evaluate $10-1+6$ as $\color{red}{3}$ following the "order of operations".

So my response is: don't ever stop being so curious, and keep learning more and more real math! But make sure to discard of those artificial substitutes (I'm trying to be nice here) that schools tend to replace the math with.

As for your example, the other answer already said it well: what happens there is that the associativity property of multiplication comes into play. Yet another reason to learn more math to see what's really going on in various calculations and such!

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$$a[b(c-d)]=ab(c-d)$$ is the way to go.

Here you are simply considering $(c-d)$ as a real number and using associative law of multiplication.