simplification when w is cube root of unity

Can you upload the original question?
Do you mean value of the expression or w^8+w^2=0 or ...

Sorry I made an error in typing

The question is: If w is the cube root of 1, find the possible values of each of the following: (the first 3 I managed to do) and then w^8 + w^10 There was no = 0 or = anything Here is the question it's form the book called Further Pure Mathematics by Brain and Mark Gaulter. I tried to get an image but it's turned out a bit too small I think

It reduces to
w + w^2
The roots of unity are
1, -1/2+/-i*sqrt(3)/2
The two complex roots are conjgates.
...

I got that it reduced to w + w^2. I have now found out a little extra (which is now obvious): if a cube root of 1 is 1 then w + w^2 2. Simple when you know how.

Thanks again, without your help I would have been struggling for even longer.

Also solving
w^3 = 1
Is factorized as
(w - 1)(w^2 + w + 1) = 0
The quadratic factor corresponds to the two complex roots and both must satisfy
w^2 + w = -1
The linear factor is
w = 1
So
w^2 + w = 2

You don't need to solve for/know the comp!ex root values, but picturing them on the unit circle (modulus 1, arg 0, 120, 240) gives insight. In particular, why w^2 is the conjugate of w.

I would like to ask another question on the same topic if you don’t mind. Question 7c from the image I attached gave just the answer to be w^2 which I got. Could a numerical answer have been one? I don’t see why not but it’s now bugging me that the book answer was not 1.

Is it
(w+w^4)/(w^2+w^5)
If so, that is
1/w = w^2
So it could (numerically) be 1,-1/2+/-isqrt(3)/2 as if w is a root of unity, so is w^2.

Thanks. I wonder why they used w^2 as an answer on this question and numbers on the others. None of the questions had the answers -1/2+/-isqrt(3)/2 but my book has explained where they come from and I understand it.