Find $w+\frac{1}{w} + \frac{1}{w^2}$ where $w^4=1$, $w\neq 1$ Let $w$ be a root of the equation $z^4 = 1$ and $w$ is not equal to $1$. 
I would like to find the value of $w+\frac{1}{w} + \frac{1}{w^2}$.
So for this question since w is a root, 
I first substituted w and got $w^4=1$
Which is $w * w^3 = 1$. 
Since $w$ is not equal to $1$, 
$w^3=1$
Which can be rewritten as $(w-1) (1 + w + w^2) = 0$
And since $w \neq 1,$ $(1 + w + w^2) = 0$
On simplifying the given equation, 
I get 
$w^3 + w + 1
= 2+w$ (Is this right approach to go about it?)
 A: Solution one:
The four roots are $1,-1,i,-i$. Simply plugging in the last three we get $-1$.
Solution two:
$z^4-1=0$ implies $(z^2-1)(z^2+1)=0$ implies $(z-1)(z+1)(z^2+1)=0$. Since we don't consider $1$, we may divide by $(z-1)$ to get: $(z+1)(z^2+1)=0$ or $z^3+z^2+z+1=0$.
Note $$\omega+\frac{1}{\omega}+\frac{1}{\omega^2}=\frac{\omega^3+\omega+1}{\omega^2}=\frac{-\omega^2}{\omega^2}=-1$$
A: 
since w is a root, I first substituted w and got w^4=1 Which is w * w^3 = 1. Since w is not equal to 1, w^3=1

This step is wrong. Compare to the obviously false: $(-1)(-1)=1$ and since $(-1) \ne 1$ then it must be that $(-1) =1$.
What you can derive from $w^4=1$ however is that $\frac{1}{w} = w^3$ and $\frac{1}{w^2} = w^2$. So the sum reduces to $w+w^2+w^3=-1$ (which holds because $1+w+w^2+w^3=\frac{1-w^4}{1-w}=0$ given $w^4=1, w \ne 1$).
A: You can't conclude $w^3=1$, but you do have the right idea of factoring $w^n -1$. Also use the fact that if you see $w^2$ you can replace it with $w^4/w^2$ if you need to.
