The equation $8x^3+4x^2-4x-1=0$ has $3$ roots $\displaystyle \cos\left(\frac{2\pi}{7}\right), \cos\left(\frac{4\pi}{7}\right)$ and $\cos\left(\frac{6\pi}{7}\right)$. Using Cardano's method and Euler's formula, I can go so far as to show that the solutions are $\displaystyle\frac{\sqrt[3]{7\sqrt{7}}}{3}\cos\left(\frac{\tan^{-1}(3\sqrt{3} )}{3}\right)-\frac{1}{6}, \displaystyle\frac{\sqrt[3]{7\sqrt{7}}}{3}\cos\left(\frac{\tan^{-1}(3\sqrt{3} )+2\pi}{3}\right)-\frac{1}{6}$ and $\displaystyle\frac{\sqrt[3]{7\sqrt{7}}}{3}\cos\left(\frac{\tan^{-1}(3\sqrt{3} )+4\pi}{3}\right)-\frac{1}{6}$. But I have no clue about how to arrive at the aforementioned solutions.
Edit: @lab_bhattacharjee, Thanks a lot. Is there a way to tell if an equation has such simplified solutions?
