<s>[INST] 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? 
 [/INST] 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, $(z- 1)(1+ z+ z^2+ z^3)= z+ z^2+ z^3+ z^4- 1- z- z^2- z^3= -1+ (z- z)+ (z^2- z^2)+ (z^3- z^3)+ z^4= z^4- 1$.
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