Htydjtk

Hope this makes sense.

$$ atimes b ÷ c $$

$$=atimesdfrac{b}{c}$$

$$=atimes btimesdfrac{1}{c}$$

$$=(atimesdfrac{1}{c})times b$$

$$=dfrac{a}{c}times b$$

$$=a÷ctimes b$$

Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$

So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.

Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.

So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.

Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.

So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.

The non-zero real numbers form an abelian (commutative) group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.

So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$

The exact same situation occurs with addition and subtraction instead of multiplication and division. Do you find it hard to justify rewriting say $a+b-c$ to $a-c+b$? And if not, which laws are you using to justify this? I would say associativity, commutativity, and the special equivalence $x-y=x+(-y)$ (which can be construed to be the definition of subtraction, although historically and in teaching subtraction is introduced before introducing negative numbers). In detail one has inserting parentheses to make explicit the one implied by convention)
$$eqalign{
a+b-c &= (a+b)-c \&= (a+b)+(-c) \&= a+(b+(-c)) \&
= a+((-c)+b) \&= (a+(-c))+b \&= (a-c)+b &= a-c+b}
$$
By perfect analogy, let us do that with '$times$' replacing '$+$', and '$div$' replacing '$-$':
$$eqalign{
atimes bdiv c &= (atimes b)div c \&= (atimes b)times (div c) \&= atimes (btimes (div c)) \&
= atimes ((div c)times b) \&= (atimes (div c))times b \&= (adiv c)times b &= adiv ctimes b}
$$
Note how I just invented the unary use of '$div$', where $div x$ of course means the multiplicative inverse of$~x$, just as $-x$ means its additive inverse. Now why did nobody think of that before? It could have avoided the ugliness of$~x^{-1}$, which has to borrow exponentiation and additive inverse to get multiplicative inverse.

Start with 56 × 100 ÷ 8.

Swap the first two factors which is definitely allowed: 100 × 56 ÷ 8.

Now put parentheses around part of it: 100 × (56 ÷ 8).

Again, swap: (56 ÷ 8) × 100.

Take off the parentheses: 56 ÷ 8 × 100.

And there you go! transforming one into the other using only multiplicative commutation (plus the fact that one of the factors commutated can be a product, not just a simple value).

Your confusion comes from a fact that is rarely spoken of in school.

The real numbers constitute a so called field where we have two operations, multiplication and addition. These two in their own create a so called group with the given set, that is the real numbers $mathbb{R}$.

This may sound really cryptic but let me clear it up. In a group you need a set $X$ and have one and only one operation $*$, you have a neutral element $e$ fullfilling that it doesnt change the element which is combined with it, all the members of the set can be combined using $*$ which will result in another element of the set $X$ and all the elements have so called inverses that is for each element $a$ in $X$ you can find another element $b$ such that $a*b=e$.

Now this is still a bit formal and cryptic so let us find an example. The whole numbers, also called as the integers $mathbb{Z}$ with usual addition forms a group.

You can add integers and you end up with integers, $0$ serves as the neutral element since any integer plus $0$ is the same integer, unchanged and you also have inverses the so called negative numbers $5 + (-5)=0$. Because of lazyness we usually write this as $5-5=0$ giving the false idea that we also have minus $"-"$ as an operation but that is WRONG.

Another exapmle could be the integers with the operation multiplication. We can multiply any integers together resulting another integer so it looks okay so far, we can find a neutral element, namely $1$ which fullfills the desired property, namely any integer times one is the same integer but we fail to find inverses. You cannot find an integer to multiply, say $5$ with to get the neutral element $1$. So this is NOT a group.

All possible fractions also callet as quotients $mathbb{Q}$ is a group with respect to multiplication (everything looks similar to what stands directly above here) but here you will find inverses. $5$ can be multiplied by $frac{1}{5}$ to get one. But once again this DOES NOT make division an operation. It is tempting to think of $frac{1}{5}$ as one divided by five which may suggest that division is an actual operation in this group but $frac{1}{5}$ is an element.

So after this short tour into abstract algebra let us consider your question again

Remember we are working in $mathbb{R}$, the field of real numbers which means that $mathbb{R}$ is a group with respect to both addition and multiplication. Or even better you could translate $frac{1}{8}$ as the inverse of $8$ with respect to multiplication

$$
frac{56}{8}cdot 100
$$
actually means that take three elements of $mathbb{R}$, namely $frac{1}{8}, 100$ and $56$ and multiply them together and since multiplication is commutative you may do this in any order. Try not to think of division as an operation rather than a way to express elements of this group.

BONUS FACT:

thinking this way also helps debunking mysteries about subtraction

$$
5-8neq 8-5
$$
which is inherently wrong since once again it suggests that minus is an operation. But using the correct way of thinking, that is the minus sign is "glued" to a number as an attribute so to say you get

$$
5+(-8)
$$

and since addition is also commutative you get

$$
(-8)+5
$$

Hope the long text will not scare you away and I could help a bit

With multiplication and division alone (no addition or subtraction) they are associative with respect to one another, BUT division itself is NOT commutative.

So basically you can do the multiplication or division in either order, BUT you must respect which way you interpret the inputs to the left or right of the operator's sign. If you flip this it has the effect of flipping the inputs which is equal precisely when the operator is commutative. As you noted, with division, it is not commutative.