$$ \sqrt[3]{\cos \frac{2\pi}{7}} + \sqrt[3]{\cos \frac{4\pi}{7}} + \sqrt[3]{\cos \frac{6\pi}{7}}$$

I found the following

  • $\large{\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}=-\dfrac{1}{2}}$

  • $\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7} + \cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7} + \cos \frac{6\pi}{7}\times\cos \frac{2\pi}{7}}=-\dfrac{1}{2}$

  • $\large{\cos \frac{2\pi}{7}\times\cos \frac{4\pi}{7}\times\cos \frac{6\pi}{7}=\dfrac{1}{8}}$

Now, by Vieta's Formula's, $\large{\cos \frac{2\pi}{7}, \cos \frac{4\pi}{7} \: \text{&} \: \cos \frac{6\pi}{7}}$ are the roots of the cubic equation


And, the problem reduces to finding the sum of cube roots of the solutions of this cubic.

For that, I thought about transforming this equation to another one whose zeroes are the cube roots of the zeroes of this cubic by making the substitution

$$x\mapsto x^3$$

and getting another equation


However, this new equation will have some extra roots too and we can't directly use Vieta's to get the desired sum.

Also, it's given that the sum evaluates to a radical of the form


where $a, b, c \: \text{&} \: d \in \mathbb Z$

Can somebody please help me with this question?

  • $\begingroup$ There is an explicit formula for the roots of a cubic, so you might try that. But is there any reason to expect a clean formula? $\endgroup$ – Thomas Andrews Dec 12 '14 at 21:12
  • $\begingroup$ @ThomasAndrews In the question, I have been given that the sum evaluates to $\sqrt[3]{\frac{1}{d}( a - b\sqrt[b]{c})}$ where $a,b,c \: \text{&} \: d \in \mathbb Z$ $\endgroup$ – user196761 Dec 12 '14 at 21:16
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    $\begingroup$ That seems like something to put in the question, then. Give us the information you have if you are seeking help. $\endgroup$ – Thomas Andrews Dec 12 '14 at 22:02
  • $\begingroup$ @ThomasAndrews Sorry, I've added it now. $\endgroup$ – user196761 Dec 12 '14 at 23:02
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    $\begingroup$ Related posts. $\endgroup$ – Lucian Dec 12 '14 at 23:17

let $$x=\sqrt[3]{\cos{\dfrac{2\pi}{7}}},y=\sqrt[3]{\cos{\dfrac{4\pi}{7}}},z=\sqrt[3]{\cos{\dfrac{6\pi}{7}}},$$ then we have $$\begin{cases} x^3+y^3+z^3=-\dfrac{1}{2}\\ (xy)^3+(yz)^3+(xz)^3=-\dfrac{1}{2}\\ (xyz)^3=\dfrac{1}{8} \end{cases}$$ use this identity $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc$$ so $$\begin{cases} (x+y+z)^3-3(x+y+z)(xy+yz+xz)+3xyz=-\dfrac{1}{2}\\ (xy+yz+xz)^3-3(xy+yz+xz)[xyz(x+y+z)]+3x^2y^2z^2=-\dfrac{1}{2}\\ xyz=\dfrac{1}{2} \end{cases}$$ let $$u=x+y+z, v=xy+yz+xz$$ then we have $$\begin{cases} u^3-3uv+2=0\\ 4v^3-6uv+5=0 \end{cases}$$ so we have $$\Longrightarrow 4v^3-2u^3+1=0, v=\dfrac{u^3+2}{3u}$$ so $$4\left(\dfrac{u^3+2}{3u}\right)^3-2u^3+1=0\Longrightarrow 4u^9-30u^6+75u^3+32=0$$ let $t=u^3$,so we have $$4t^3-30t^2+75t+32=0$$ let $t=\dfrac{5}{2}-a$,then $$4\left(\dfrac{5}{2}-a\right)^3-30\left(\dfrac{5}{2}-a\right)^2+75\left(\dfrac{5}{2}-a\right)+32=0$$ $$\Longrightarrow 4a^3=\dfrac{189}{2}\Longrightarrow a=\dfrac{3\sqrt[3]{7}}{2}$$ so $$u=x+y+z=\sqrt[3]{\cos{\dfrac{2\pi}{7}}}+\sqrt[3]{\cos{\dfrac{4\pi}{7}}}+\sqrt[3]{\cos{\dfrac{6\pi}{7}}}=\sqrt[3]{\dfrac{1}{2}\left(5-3\sqrt[3]{7}\right)}$$

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  • $\begingroup$ Whoa! Thanks a lot!! $\endgroup$ – user196761 Dec 13 '14 at 14:47
  • $\begingroup$ @math110: Can you kindly look at this related question? $\endgroup$ – Tito Piezas III Dec 14 '14 at 20:22

Try using this: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ Let $a^3= \cos{\frac{2\pi}{7}},b^3= \cos{\frac{4\pi}{7}},c^3= \cos{\frac{6\pi}{7}} $.

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  • $\begingroup$ I'm sorry, but I still can't figure out what to do with the second factor in the $\text{R.H.S.}$ $\endgroup$ – user196761 Dec 12 '14 at 22:52

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