Disjunctive normal form expansion I do not understand this at all.

Find the sum-of-products expansions of these Boolean functions.

*

*$F(x, y, z) = x + y + z$


*$F(x, y, z) = (x + z)y$


*$F(x, y, z) = x$


*$F(x, y, z) = x y$

How is $x$ not just $x$? This makes no sense to me at all and my book wants me to memorize about 20 laws and stuff to transform these to some form that I don't understand. I don't understand the point of this, I don't understand the goal of this and I don't understand the procedure. I can't even find an explanation online that makes sense to me.
What am I supposed to do?
$x+y+z$ looks like a sum of products to me already. What is wrong with it?
Why isn't $(x+y)z$ just $xz + zy$? My book says that is wrong.
Books answers:


*

*$xyz + xy\bar z + x \bar yz + x\bar y \bar z + \bar x yz + \bar x y \bar z + \bar x\bar y z$
*$ xyz + xy \bar z + \bar xyz$

*

*$xyz + xy \bar z + x \bar y z +  x \bar y\bar z $


*$ x\bar yz + x\bar y \bar z$

 A: Each of the above except for $(x + y) z$ is already expressed as the sum-of-products (SOP). $$(x + y)z = xz + yz = xz + zy$$
You are correct that $F(x, y, z) = x$ is already in disjunctive normal form (SOP form). Perhaps your book wants you to write $F(x, y, z) = x = 1x + 0y + 0z$, but I find that a bit pedantic!
In that spirit, $F(x, y, z) = x + y + z = 1x + 1y + 1z$, and $F(x, y, z) = xy = xy + 0z$. But by most authoritative sources on the matter, these alternatives are unnecessary, and all but $F(x, y, z) = (x + y)z$ are perfectly respectable ways to express the functions as a sum-of-pr0ducts.
A: I think the goal of the answers your book is using is to guarantee that there is exactly one term which is 1 when the whole thing is 1. One way to do this is to expand the functions into essentially glorified truth tables, and this is what appears to be happening. You can see that every term has three factors, so you're just trying to figure out how to modify your intuitive answers (like $x+y+z$) into something that will do this.
A little memorization would probably be good, but you can always logic one of these out, like so: Consider $(x+y)z$. We can still use normal algebra rules, so this is the same as $xz+yz$. That looks a lot like an "intuitive SoP" so that's the first big step.
The first term doesn't care about $y$ so we could write it as $xzy+xz\bar y$. The second term doesn't care about $x$ so we could write it as $yzx+yz\bar x$. So the whole thing becomes $xyz+x\bar yz + xyz + \bar xyz$. But $xyz$ is in there twice, so we can take it out. That leaves us with $xyz+x\bar yz+\bar xyz$.
You can probably tease out the general algorithm at play: for any term not including $x_i$, multiply it by $(x_i+\bar{x_i})$. Do this until every term in the expanded version has all variables, and then remove duplicates.
[Personally I think this is hideous and kind of ruins the point of using function notation in the first place, but… you have to play along with the rules sometimes. Plus, in any case it's probably good to at least know that such a thing can be done.]
