Dividing rational expression? Just to clarify, I do not want anyone to do the problem for me.  I just need someone to explain this stuff to me because I'm completely lost, and I would really appreciate it if you could take the time to do so.
In the book I've been reading they have this example:
$$\frac{y^3-1}{y^2-1} \div \frac{y^2 + y + 1}{y^2 + 2y + 1}$$
So I understand the basic concept: you flip the fraction so that you're multiplying, then factor everything on top and everything on the bottom, then you cancel things that are the same on top and on the bottom. What I don't understand is, when they factor the numerator, they get rid of a "2" and I'm not really sure how they did that: 
$$\frac{y^3 - 1}{y^2 - 1} \div \frac{y^2 + 2y + 1}{y^2 + y + 1}$$
Factor:
$$(y - 1)(y + 1)(y + 1)(y^2 + y + 1) \div (y + 1)(y - 1)(y^2 + y + 1)$$
Cancel things out:
$$y + 1$$
So the problem I have is that they have $(y^3 - 1)(y^2 + 2y + 1)$ on top, but somehow they factor it out to: $(y - 1)(y + 1)(y + 1)(y^2 + y + 1)$; how is that possible? I get that $(y^3 - 1)$ factors out to:
$$(y - 1)(y + 1)(y + 1)$$
but how did they get rid of the "2" in $(y^2 + 2y + 1)$ so that it's $(y^2 + y + 1)$?
 A: The $y^2+2y+1$ is still there as $(y+1)(y+1)$.  The factor $y^2+y+1$ is from the denominator of the last fraction, which, after inverting and multiplying, is on top.
A: You have to use the factorisation of $y^3-1$ and $y^2+2y+1$:
$$\dfrac{y^{3}-1}{y^{2}-1}\times\dfrac{y^{2}+2y+1}{y^{2}+y+1}=\frac{( y-1)(y^2+y+1)}{(y-1)(y+1)}\times\frac{(y+1)^2}{y^2+y+1}=y+1.$$
A: $(y^3-1)/(y^2-1) ÷ (y^2 + y + 1)/(y^2 + 2y + 1)=$
$(y^3-1)/(y^2-1)\times (y^2 +2 y + 1)/(y^2 + y + 1)=$
$\frac {(y^3 -1)(y^2+2y+1)}{(y^2-1)(y^2+y+1)}=M$
I think you've followed and figured all that on your own.
Now comes.... experience, intuition and elbow grease.
Does $y^3 - 1$ factor?  I know it does because I've been doing this stuff for 35 years.  If it does then it will be of the form $(ay^2 + by + c)(dy + e)$ where $ad = 1$ and $ce = -1$ and ... some mess of inside stuff will cancel out.  (Actually $ady^3 + (bd + ea)y^2 + (be+ cd)y + ec = x^3 - 1$ so $ad = 1; bd+ea = 0; be+cd=0; ec=-1$).  
First guess is $e =\pm 1; c= \mp1; a = 1;d = 1$ and we get $(y^2 + by \mp 1)(y \pm 1) = y^3 + by^2 \mp y \pm y^2 \pm by - 1= y^3 +(b \pm 1)y^2 + (\pm b \mp 1)y  -1 = y^3 - 1$ so $b \pm 1 = 0$ and $b - 1 = 0$ so $b = 1$ and $y^3 - 1 = (y-1)(y^2 + y + 1)$.  
Indeed, we should probably learn and put it in a arsenel of tricks that $(x-1)(x^n + x^{n-1} + ...... + x + 1) = x^{n+1} + (1-1)x^{n-1} + ..... + (1-1)x -1 = x^{n+1} - 1$ is a good bit of factoring that we will use a lot in the future.
So $y^3 - 1 = (y-1)(y^2 + y+ 1)$ and we have:
$M =\frac{(y^3 - 1)(y^2+2y+1)}{(y^2 - 1)(y^2+ y + 1)} = \frac{(y-1)(y^2+y+1)(y^2 + 2y + 1)}{(y^2- 1)(y^2 + y+1)} = \frac{(y-1)(y^2 + 2y + 1)}{y^2 - 1}$
Now does $y^2 - 1$ factor.  Well, above we say $(x-1)(x^n + .... + 1) = x^{n+1} -1$ so $y^2 - 1 = (y-1)(y+1)$.  We can check that: $(y-1)(y+1) = y^2 - y + y - 1 = y^2-1$.
So
$M = \frac{(y-1)(y^2 + 2y + 1)}{y^2 - 1}=$
$\frac{(y-1)(y^2 + 2y + 1)}{(y-1)(y+1)} =\frac{(y^2 + 2y + 1)}{(y+1)}$ 
Now the question is: does $y^2 + 2y + 1$ factor?  Well, yes... $y^2 + 2y + 1 = (y+1)^2$
So $M = \frac{y^2 + 2y + 1}{y+1} = \frac{(y+1)^2}{y+1} = y+1$. and.... we are done.
A: We have: $\dfrac{\dfrac{y^{3}-1}{y^{2}-1}}{\dfrac{y^{2}+y+1}{y^{2}+2y+1}}$
You were correct in stating that we "flip the fraction", but you incorrectly displayed a division symbol rather than a multiplication symbol.
The author never got rid of the $2y$; the expression $y^{2}+2y+1$ was factored into $(y+1)(y+1)$.
$=\dfrac{y^{3}-1}{y^{2}-1}\times\dfrac{y^{2}+2y+1}{y^{2}+y+1}$
$=\dfrac{(y-1)(y^{2}+y+1)}{(y+1)(y-1)}\cdot\dfrac{(y+1)(y+1)}{(y^{2}+y+1)}$
$=y+1$
