<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] Note $(x^2-x^3)(x^4-x)=-x^7+x^6+x^4-x^3$. The key is that $x$ is a root of unity, particularly one that gives $x^5=1$. This means $x^6=x$ and $x^7=x^2$ while $x^4=x^*$ and $x^3=(x^2)^*$ which means that our imaginary components fall out since $z+z^*=2a$ for $z=a+bi$. Squaring both sides to get rid of the radical  reduces the problem nicely.
sorry for any errors as I've posted this from my phone.
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