Quick Proof questions

I assume you mean 1 + w + w^2 = 0?

The sum of the roots is acceptable I believe, but only if you have proved that 1, w and w^2 are the roots already. (which you did by saying what the 3 roots are, but you might have to do a little more to show that they are in fact the 3 roots of the equation)

Another way is to factorise z^3-1=0
into (z-1)(z^2+z+1) = 0 (you can figure this out the first time through polynomial division and then I think you can quote the result, at least you can in my system)
But w is non-real so w=/=1 so 1 + w + w^2 = 0

Ah yea, After finding the 3 roots I would let one non-real root be $w$ and then show that $w^6 = (w^2)^3 = 1^6 = 1$ and so $w^2$ is also a root and then proceed as I did above.

I probably wouldn't have spotted your method and it would take too long for me to divide the thing by $z-1$

I recommend you ask your teacher whether you are allowed to quote the general result:

z^n -1 = (z-1)(z^(n-1) + z^(n-2) + z^(n-3) + ... + 1)

rather than actually deriving it each time. If you are allowed to, it is a slight bit quicker than showing that the roots are 1, w and w^2 and then using the sum of the roots. If not, your method is perfectly valid.

Self teaching FM lol, that's why I post about 3-4 questions in here a day.