An operation on a whole is equal to an operation on each part? I've been pondering this question as to how and when you can perform an operation on a complete "unit" and the answer is the same when performing the operation on the individual parts of the "unit"
For example:
$$
\sqrt{25\div4} = \sqrt{25}\div\sqrt{4}.
$$
We took the square root operation and applied it to $25$ and $4$ separately to get the same result. 
However this is not the case for:
$$
(4+1)^2 \neq 4^2 + 1^2
$$
However this is true again for:
$$
\frac{2(7-6)}{2^2} = \frac{2(7)}{2^2} - \frac{2(6)}{2^2}.
$$
We took multiplication by $2$ and division by $2^2$ and applied them to $7$ and $6$ individually.
My question is: Can you always distribute operations to the parts of a unit and get the same result as when you would perform the operation on the whole unit.  
 A: As has been hinted in the comments, there is a deeper underlying structure here. But since you haven't been introduced to abstract algebra yet, I won't go down that road.
What I can say is that this does not hold for all operations. I would give a counter example, but you already gave one:
$$(4+1)^2\neq4^2+1^2.\tag{1}$$
There is also no general rule for when it does hold. One of the reasons is that there is no definition of what a part of an expression is. For instance, in $(1)$ you consider "$4$" and "$1$" to  be different part, but in your first example you also consider "$25$" and "$ 4$" to be differnt parts of ${25}\div4$. But in this case we do have that 
$$(25\div 4)^2=25^2\div4^2.\tag{2}$$
But when the operation is "taking powers" (I don't know if you learned this already), then we have:
$$2^{25\div 4}\neq2^{25}\div2^{4}.\tag{3}$$
You wanted to know if we can

distribute operations to the parts of a problem and get the same result as performing an operation on a whole unit.

Examples $(1)$, $(2)$ and $(3)$ show us that this depends on


*

*what the operation is

*what we consider to be the "parts" of a unit.


The answer to your question will be different for each opereation and for each definition of "a part". 
So you have to figure this out each time you learn about a new operation.
