1Sorry, but i cant continue, becouse i don't know sr.
:/
\sqrt[3]{cos(\frac{2\pi}{7})}+\sqrt[3]{cos(\frac{4\pi}{7})}+\sqrt[3]{cos(\frac{6\pi}{7})}
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3 Answers
39i have done such questions without cube root
but with cube root am not sure
any hints ?
1The roots of x7 - 1=0 are cos2kÎ 7 + isin2kÎ 7 , k = 0,1,2....6
hence roots of x6+x5+...+x+1 = are
xk = 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 xk + xk (bar) = 2cos2kÎ 7
so the given expressions are the roots of the cubic....
Now proceed