<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] Replacing the earlier sequence of hints with a solution now. The old aswer is scratched.
We know that $0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. As $x-1\neq0$ this implies that
$$1+x+x^2+x^3+x^4=0.\qquad(*)$$
From $x^5=1$ we also deduce that $x^4=x^{-1}$. This allows a rewrite:
$$
S=(x^2-x^3)(x^4-x)=(x^2-x^3)(x^{-1}-x)=(x-x^2)(1-x^2)=x-x^2-x^3+x^4.
$$
Let's square this. We get
$$
\begin{aligned}
S^2&=(x^2+x^4+x^6+x^8)\\
&+2(-x^3-x^4+x^5+x^5-x^6-x^7).\qquad(**)
\end{aligned}
$$
Because $x^8=x^5\cdot x^3=x^3$ and $x^6=x$, the first term above is
$$
(x^2+x^4+x^6+x^8)=x^2+x^4+x+x^3=-1,
$$
by equation $(*)$. The latter term in parens is similarly simplified to
$$
(-x^3-x^4+x^5+x^5-x^6-x^7)=2-(x^3+x^4+x+x^2)=3.
$$
Plugging these both into $(**)$ gives
$$
S^2=5.
$$
So we know that $S=\pm\sqrt5$, and the remaining task is to determine the sign.
From a picture of the unit circle in the complex plane, we see that all the terms in the r.h.s. of
$$
S=x-x^2-x^3+x^4
$$
have positive real parts. Therefore $S=\sqrt5$.

No trigonometry needed. Just $(*)$ and properties of roots of unity. Study Gauss sums for generalizations to primes $>5$.
</s>