Find de value

The roots of x⁷ - 1=0 are cos2kÎ 7 + isin2kÎ 7 , k = 0,1,2....6

hence roots of x⁶+x⁵+...+x+1 = are

x_k = cos2kÎ 7 + isin2kÎ 7 , k = 1,2...6

Put x + 1x = y

the eqn of sixth degree can be written as

( x³ + 1x³ ) + ( x² + 1x² ) + ( x + 1x ) + 1 = 0

Clearly the rots are conjugate

so x_k + x_k (bar) = 2cos2kÎ 7

so the given expressions are the roots of the cubic....

Now proceed

Sorry, but i cant continue, becouse i don't know sr.
:/