I'm working my way through Mathematical Methods in the Physical Sciences and came across the following problem:
Use a computer to find the three solutions of the equation $x^3−3x−1=0$. Find away to show that the solutions can be written as $2 cos(\pi/9)$, $-2cos(2\pi/9)$, $-2cos(4\pi/9)$.
I can easily find the roots numerically and subsequently check that they do, in fact, match the expressions given, but I'm at a loss as to how find the roots analytically.
Based on the section that this problem is from, I suspect that I should replace $x$ with a complex exponential, say $e^{iy}$, and use De Moivre's formula. However, no amount of algebra has gotten me any closer to even one of these solutions.
Any input on whether this is correct thinking or how to go about this problem would be greatly appreciated.
Consider the general cubic equation $$x^3+px+q=0,$$and let $$x=2a\cos t=a(e^{it}+e^{-it}).$$ Developing, you get $$x^3+px+q=a^3(e^{3it}+3e^{it}+3e^{-it}+e^{-3it})+ap(e^{it}+e^{-it})+q=0.$$ If you make $$3a^3+ap=0,$$ all terms in $e^{\pm it}$ cancel out and you are left with $$a^3(e^{3it}+e^{-3it})+q=0,$$i.e. $$\cos3t=-\frac q{2a^3},$$ where $$a=\sqrt{-\frac p3}.$$ In the given case, $a=1$, $\cos3t=\frac12$, $3t=\pm\frac\pi3+2k\pi$, $t=\frac{\pm1+6k}9\pi$.