<s>[INST] On the roots of $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$ The Problem Prove that $\tan \frac{\pi}{15}$ is a root of the equation $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$, and find the other roots.
Source: Question 17 from the complex numbers chapter from Bostock's Further Pure Mathematics
Thoughts Using the identity $\tan 5\theta =\frac{5\tan\theta-10\tan^3\theta-\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$, I can show that $\tan \frac{\pi}{15}$ is a root of $t^5+5\sqrt3t^4+10t^3-10\sqrt3t^2-5t+\sqrt3=0$, but it isn't clear how to proceed. Alternatively, I can derive and use a formula for $\tan 4\theta$ to obtain a quartic equation as desired, but that will require me to find $\tan \frac{4\pi}{15}$ which I cannot.
I suspect this is not difficult though I can't see how to proceed. Hints would be appreciated.
 [/INST] First of all, $\tan5\theta=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$
Set $\tan5\theta=\sqrt3=\tan\frac\pi3$
$\implies 5\theta=n\pi+\frac\pi3=\frac{(3n+1)\pi}5$ where $n$ is any integer
So, the values of $\theta$ can be set from $n=0,1,2,3,4$
Observe that for $n=3,\tan\theta=\tan\frac{2\pi}3=\tan\left(\pi-\frac\pi3\right)=-\tan\frac\pi3=-\sqrt3$
So rearrange $\displaystyle\sqrt3=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$ and divide by $\tan\theta -(-\sqrt3)=\tan\theta+\sqrt3$ to get the equation with roots $\displaystyle\tan\frac{(3n+1)\pi}{15}$  where $n=0,1,2,4$
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