Expanding brackets problem: $(z - 1)(1 + z + z^2 + z^3)$ I have:
$(z - 1)(1 + z + z^2 + z^3)$
As, I have tried my own methods and enlisted the help on online software, but as well as them not all arriving at the same solution, I can't follow their reasoning.
I tried to gather all the like terms:
$(z - 1)(z^6+1)$
And then thought it may offer a route to a difference or two cubes if I shifted powers across the brackets, but that's $z^7$ in total, isn't it? I can't just do $z^7-1$, can I?
If I try to expand, I see:
$(z - 1)(z^6 + 1) = z^7+z-z^6-1$
Well...
$z^7-z^6=z$
Add the other z leaves:
$z^2-1$?
I expect this is all wrong.
Will you help please?
On a side note, I have difficulty with understanding how I'm expected to tag this post accurately when I can't used "expand", "multiply", or use any of the subjects it covers as tags? 
 A: You seem to have some very unfortunate ideas about algebra!  As David Mitra said, When you are adding powers of z you do not add the powers themselves.  That is a property of multiplication: $z^n\cdot z^m= z^{n+ m}$.
To multiply $(z- 1)(1+ z+ z^2+ z^3)$ use the "distributive law" $a(b+ c)= ab+ ac$ and $(b+ c)a= ab+ ac$.
Think of $z- 1$ as $(b+ c)$ with $b= z$ and $c= -1$ and a as $(1+ z+ z^2+ z^3)$.
$(z- 1)(1+ z+ z^2+ z^3)= z(1+ z+ z^2+ z^3)- 1(1+ z+ z^2+ z^3)$.  
Now, for each of those, use the distributive law again: $z(1+ z+ z^2+ z^3)= z(1)+ z(z)+ z(z^2)+ z(z^4)$. 
NOW use the rule for adding exponents (with $1= z^0$ and $z= z^1$): $z(1)+ z(z)+ z(z^2)+ z(z^3)= z+ z^2+ z^3+ z^4$.  
And, of course, $-1(1+ z+ z^2+ z^3)= -1- z- z^2- z^3$.   
Putting those together, $(z- 1)(1+ z+ z^2+ z^3)= z+ z^2+ z^3+ z^4- 1- z- z^2- z^3= -1+ (z- z)+ (z^2- z^2)+ (z^3- z^3)+ z^4= z^4- 1$.
A: I like to do factorizations like this by writing 
$$
z^7-1=(z-1)(\cdots)
$$
and try to figure out what should replace the dots.  First, we'd like to have a factor of $z^7$ in the end result, so the $z$ on the right side should be multiplied by $z^6$ to get this.  Then, we write
$$
z^7-1=(z-1)(z^6+\cdots).
$$
At this point, by the distributive law, the right side will multiply out to $z^7-z^6+\cdots$.  Since we don't want the $z^6$ in the final product, it must cancel with something.  We can cancel the $-z^6$ by multiplying $z$ by $z^5$ to get
$$
z^7-1=(z-1)(z^6+z^5+\cdots)
$$
because after distributing, the product is $z^7-z^6+z^6-z^5+\cdots$ and the $z^6$'s cancel.  Now, we're left with $z^7-z^5+\cdots$, so the $-z^5$ must cancel with another term, since $z^5=z\cdot z^4$, we have
$$
z^7-1=(z-1)(z^6+z^5+z^4+\cdots).
$$
At this point, the product is $z^7-z^4+\cdots$.  In order to cancel the $-z^4$, we introduce a $z^3$, which results in an extra $-z^3$.  This extra term is cancelled with a $z^2$, which results in an extra $-z^2$ in the product.  This $-z^2$ is cancelled with a $z$, but that gives an extra $-z$ in the product.  The extra $-z$ is cancelled with a $-1$.  Therefore, we have
$$
z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1+\cdots).
$$
But, after multiplying everything out, we find that the left side already matches the right side and there is nothing more in the dots!  Therefore,
$$
z^7=(z-1)(z^6+z^5+z^4+z^3+z^2+z^1+1).
$$
A: First off, user247327 gives an excellent answer. I don't have the reputation to comment on his comment, so here is an explanation of a critical fact you are missing. Q: why doesn't $$(1+z+z^2+z^3)=(z^6+1)?$$
Answer: When we write $z^3$, we mean $z*z*z$. Likewise $z^2=z*z$. Therefore:
$$z^3*z^2=(z*z*z)*(z*z)$$
Well, how to we write this compactly? Simple! count the number of times we multiply z. Thus we get:
$$z^3*z^2=(z*z*z)*(z*z)=z^5$$
This is why it is true in general that: $$z^a*z^b=z^{a+b}$$
If you ever forget this, just google "exponent rules", and you can see all the neat tricks. HOWEVER, there is no identity I can think of for simplifying $z^a+z^b$. 
Edit: PS, in your comment you state $z^3+z^2=z^5$. Substitute z=2 to see that this is wrong. Note $2^5=32$, but: $$2^3+2^2=8+4=12 \neq 32$$
This one counterexample is enough to show that $z^a+z^b \neq z^{a+b}$ 
A: First remove the left brackets and distribute the left terms on the right brackets:
$$(z - 1)(1 + z + z^2 + z^3)=z(1 + z + z^2 + z^3)-(1 + z + z^2 + z^3).$$
Then remove the remaining brackets and distribute what needs to be:
$$z(1 + z + z^2 + z^3)-(1 + z + z^2 + z^3)=z+z^2+z^3+z^4-1-z-z^2-z^3.$$
Finally, simplify:
$$z+z^2+z^3+z^4-1-z-z^2-z^3=z^4-1.$$
