Let $x=2\cos \theta .$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=69827

Called $x_{1},x_{2},x_{3}$ are roots of $P$. We need find $x_{1}^{5}+x_{2}^{5}+x_{3}^{5}\; ,x_{1}^{5}x_{2}^{5}+x_{2}^{5}x_{3}^{5}+x_{3}^{5}x_{1}^{5}$ and $x_{1}^{5}x_{2}^{5}x_{3}^{5}$. This is easy!

Wrong message ! See the following ...

N.T.TUAN wrote: Let $P (x) = x^{3}-3x+1.$ Find the polynomial $Q$ whose roots are the fifth powers of the roots of $P$. Proof. $\left\{\begin{array}{c}x^{3}=3x-1\\\ y=x^{5}\end{array}\right\|$ $\Longrightarrow$ $y=x^{2}(3x-1)=3(3x-1)-x^{2}$ $\Longrightarrow$ $\left\{\begin{array}{c}x^{3}-3x+1=0\\\ x^{2}-9x+(3+y)=0\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}x^{2}-9x+(3+y)=0\\\ 9x^{2}-(6+y)x+1=0\end{array}\right\|$ $\Longrightarrow$ $\left|\begin{array}{cc}1 & 3+y\\\ 9 & 1\end{array}\right|^{2}=\left|\begin{array}{cc}1 & 9\\\ 9 & 6+y\end{array}\right|\cdot \left|\begin{array}{cc}9 & 3+y\\\ 6+y & 1\end{array}\right|$ $\Longrightarrow$ $(9y+26)^{2}+(y-75)\left(y^{2}+9y+9\right)=0$ $\Longrightarrow$ $\boxed{y^{3}+15y^{2}-198y+1=0}$.

Here is an interesting solution using tirogonometry: We now that, $\sin{3\phi}=3\sin{\phi}-4\sin^{3}{\phi}$ now lets multipliy both sides by $2$ and rearrange it: $(2\sin{\phi})^{3}-3(2\sin \phi)+2\sin 3\phi=0$ $(1)$ now this looks quite similar to the original one: ($x^{3}-3x+1$) We can read this as , if every time that for $\phi$ the last part of $(1)$ is equal to the last part of $P$ then for this angle $2\sin \phi$ is the solution for $(1)$. So : $2\sin 3\phi =1$ The solution for this equation: $\phi =10^{\circ}+k\cdot 120^{\circ}$ and $\phi = 50^{\circ}+k \cdot 120^{\circ}$ using that the sine function is symmetrink we get three differant sine values, that means three differant solutions: The solutions for $(1)$ are: $2\sin 10^{\circ}$ $2\sin 130^{\circ}$ $2\sin 250^{\circ}$ So if the $Q$ polynomial has the following form: $Q=x^{3}+ax^{2}+bx+c$ Then: $a=-2^{5}(\sin^{5}10^{\circ}+\sin^{5}130^{\circ}+\sin^{5}250^{\circ})=15$ $b=2^{10}(\sin^{5}10^{\circ}\cdot \sin^{5}130^{\circ}+\sin^{5}130^{\circ}\cdot \sin^{5}250^{\circ}+\sin^{5}250^{\circ}\cdot \sin^{5}10^{\circ})=-198$ $c=-2^{15}\cdot \sin^{5}10^{\circ}\cdot \sin^{5}130^{\circ}\cdot \sin^{5}250^{\circ}=1$ So the polynomial we are looking for: $Q=x^{3}+15x^{2}-198x+1$