Primes $a$, $b$, $c$, $d$ satisfying $(\cos(2\pi/7))^{1/3}+(\cos(4\pi/7))^{1/3}+(\cos(6\pi/7))^{1/3}=\left(\frac{a-b\sqrt[3]c}d\right)^{1/3}$

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Source: Enumeration 2022 Prelims conducted by the Indian Institute of Science (question $10$ under objectives)

Problem statement:

Primes $a$, $b$, $c$, $d$ satisfy the following equation. $$\left(\cos\frac{2\pi}{7}\right)^{\frac 13}+\left(\cos\frac{4\pi}{7}\right)^{\frac 13}+\left(\cos\frac{6\pi}{7}\right)^{\frac 13}=\left(\frac{a-b\times\sqrt[3]c}{d}\right)^{\frac 13}$$ Find the value of $a+b+c+d$.

My attempt:

Let $\alpha=\cos\frac{2\pi}7$, $\beta=\cos\frac{4\pi}7$, $\gamma=\cos\frac{6\pi}7$. It can be established that:

  • $\alpha+\beta+\gamma=-\frac{1}{2}$
  • $\alpha\beta+\beta\gamma+\gamma\alpha=-\frac{1}2$
  • $\alpha\beta\gamma=\frac{1}{8}$

Thus, $\alpha$, $\beta$, $\gamma$ are roots of the cubic equation $8x^3+4x^2-4x-1=0$.


Is there any way to find the sum of cube roots of the roots of a cubic? I am not sure what to do next. How do I manage such questions in a short time in the setting of a contest? I am not sure if the latter part of my question is in the scope of this site.

I would appreciate alternate ways to approach this problem as well.