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De Moivre's theorem: multiple-angle identities and cubic equations

mathematical

This question links de Moivre's theorem to solving a cubic.

Facts you may need

$(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta$ — De Moivre with $n = 3$.
$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ — Binomial expansion.
Sum and product of the roots of $ax^3 + bx^2 + cx + d$ : $-\frac ba$ and $-\frac da$ — Vieta's formulas.

Use de Moivre's theorem to show that

\cos 3\theta = 4\cos^3\theta - 3\cos\theta

Hence find the exact solutions of

8x^3 - 6x - 1 = 0

Deduce the exact values of

\cos\frac{\pi}{9} + \cos\frac{5\pi}{9} + \cos\frac{7\pi}{9}

and

\cos\frac{\pi}{9}\cos\frac{5\pi}{9}\cos\frac{7\pi}{9}