<s>[INST] Prove that $(x^2-x^3)(x^4-x) = \sqrt{5}$, where $x= \cos(2\pi/5)+i\sin(2\pi/5)$ 
Prove $(x^2-x^3)(x^4-x) = \sqrt{5}$ if $x= \cos(2\pi/5)+i\sin(2\pi/5)$.

I have tried it by substituting $x = \exp(2i\pi/5)$
but it is getting complicated.
 [/INST] $x^4+x^3+x^2+x+1=0$. Divide by $x^2$, and substitute $z=x+x^{-1}$, resulting in $z^2+z-1=0$. This has two solutions: $z_1=2cos72^\circ=\dfrac{-1+\sqrt5}{2}$, and $z_2=2cos144^\circ=\dfrac{-1-\sqrt5}{2}$.
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