Formal Derivative and Multiple Roots

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Primitive roots of 1 over a finite field

Homework Statement



The polynomial x3 − 2 has no roots in F7 and is therefore irreducible in F7[x]. Adjoin a root β to make the field F := F7(β), which will be of degree 3 over F7 and therefore of size 343. The
multiplicative group F× is of order 2 × 9 × 19. Show:
(1) that −1 is a primitive square root of 1 in F;
(2) that β is a primitive 9th root of 1 in F;
(3) that −1 + β is a primitive 19th root of 1 in F;
(4) and then that β − β2 is a primitive root in F, that is, a generator of the multiplicative group F×.

Homework Equations





The Attempt at a Solution



I am totally confused about how to even start this proof. Any help would be greatly appreciated please!
 
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β3-2=0
or, equivalent, β3+5=0
You can add and multiply polynomials with β as usual, just consider everything mod 7, and use that equation to keep the degree of your polynomials smaller than 3.

Therefore, -1 = 6 (as 6+1=0), and 6*6=...
You can calculate β, β^2 (well, nothing to do for those), β^3, β^4, ... and show that it is a primitive 9th root in that way.
(3) would be possible like that, too, but there might be a more elegant way.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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