If a is a non real root of $x^7 = 1$, find the equation whose roots are $ a + a^6 , a^2 + a^5, a^3 + a^4$ If a is a non real root of $ x^7 = 1$, find the equation whose roots are $a + a^6 , a^2 + a^5, a^3 + a^4$. This is one of the questions I have encountered while preparing for pre rmo. I feel the question requires the concept of the nth roots of unity and de moivre's theorem. But i actually couldnt work it out. Any help will be appreciated. Thanks in advance.
 A: Think of the roots as $a+a^{-1}$, $a^2+a^{-2}$ and $a^3+a^{-3}$. Then $a$ satisfies $a^6+a^5+a^4+a^3+a^2+a+1=0$ or equivalently,
$a^3+a^2+a+1+a^{-1}+a^{-2}+a^{-3}=0$. Can you write the expression
$x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3}$ as a linear combination of $(x+x^{-1})^3$, $(x+x^{-1})^2$, $x+x^{-1}$ and $1$? Putting in $a$ for $x$
would then give you an equation satisfied by $a+a^{-1}$.
A: WLOG let $a=e^{\frac{2i\pi}{7}}$. Then expand 
$$(x-(a+a^6))(x-(a^2+a^5))(x-(a^3+a^4))$$
using the facts that $a^7=1$ and $a^6+a^5+a^4+a^3+a^2+a+1=0$. 
A: Like factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$,
we can prove $a^r+a^{7-r}=a^r+a^{-r}=2\cos\dfrac{2r\pi}7$ for $r=1,2,3$
Now if $7x=2r\pi,$
$\cos4x=\cos(2r\pi-3x)=\cos3x\  \ \ \ (1) $
So, the equation whose roots are $\cos\dfrac{2r\pi}7$ for $r=0,1,2,3$
$$8c^4-8c^2+1=4c^3-3c$$
But $(1)\implies4x=2m\pi\pm3x$ where $m$ is any integer
$4x=2m\pi+3x\implies x=2m\pi,\cos x=1$
So, the equation whose roots are $\cos\dfrac{2r\pi}7$ for $r=1,2,3$
$$\dfrac{8c^4-4c^3-8c^2+3c+1}{c-1}=0$$
Can you take it from here? 
