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.
 A: 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 $>5$.
A: 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.
A: As $x=\cos\frac{2\pi}5+i\sin\frac{2\pi}5$
Using  de Moivre's formula for positive integer $n$
$$x^n=\left(\cos\frac{2\pi}5+i\sin\frac{2\pi}5\right)^n=\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5$$
$$\implies x^5=\cos2\pi=1\text{ and }x^{-n}=\frac1{x^n}=\frac1{\cos\frac{2n\pi}5+i\sin \frac{2n\pi}5}=\cos\frac{2n\pi}5-i\sin \frac{2n\pi}5$$
$$\implies  x^n+x^{-n}=2\cos\frac{2n\pi}5$$
As $x^5=1,x^6=x,x^4=x^{-1},x^7=x^2,x^3=x^{-2}$
$$\implies (x^2-x^3)(x^4-x)=x^6+x^4-x^7-x^3=x+\frac1x-\left(x^2+\frac1{x^2}\right)$$
$$\implies (x^2-x^3)(x^4-x)=2\cos\frac{2\pi}5-2\cos\frac{4\pi}5=2\cos\frac{2\pi}5-2\cos(\pi-\frac\pi5)=2\cos\frac{2\pi}5+2\cos\frac\pi5>0\ \ \ \ (0)$$
Now if $y^5=1$
Using $n$th root of unity, $\displaystyle y=\cos\frac{2r\pi}5+i\sin \frac{2r\pi}5$ where $r=0,1,2,3,4$
$r=0\implies y=1\implies$  the roots of $\displaystyle \frac{y^5-1}{y-1}=0\iff y^4+y^3+y^2+y+1=0\ \ \ \ (1)$ are $\cos\frac{2r\pi}5+i\sin \frac{2r\pi}5$ where $r=1,2,3,4$
Observe that the last equation is Reciprocal Equation of the First type like this
So, divide either sides by $y^2,$  $$y^2+y+1+\frac1y+\frac1{y^2}=0 \implies \left(y+\frac1y\right)^2+\left(y+\frac1y\right)-1=0\ \ \ \ (2)$$
We have $y+\frac1y=2\cos\frac{2r\pi}5$
and as $\cos\frac{2(5-r)\pi}5=\cos(2\pi-\frac{2r\pi}5)=\cos \frac{2r\pi}5$
the roots of $(2)$ are 
$2\cos\frac{8\pi}5=\cos\frac{2\pi}5>0$ and $\cos\frac{6\pi}5=\cos(2\pi-\frac{2r\pi}5)=\cos\frac{4\pi}5=-\cos\frac\pi5<0$
Now, $2\cos\frac{2r\pi}5=y+\frac1y=\frac{-1\pm\sqrt5}2$ where $r=(1$ or $5-1=4)$ and $r=(2$ or $5-2=3)$
So, using  Vieta's formulas,  $2\cos\frac{2\pi}5+\left(-2\cos\frac\pi5\right)=-1$ and $2\cos\frac{2\pi}5\left(-2\cos\frac\pi5\right)=-1$
$$\implies 2\cos\frac{2\pi}5+2\cos\frac\pi5=\sqrt{\left(2\cos\frac{2\pi}5-2\cos\frac\pi5\right)^2+4\cdot2\cos\frac{2\pi}5\cdot2\cos\frac\pi5}=\sqrt{(-1)^2+4\cdot1}=\sqrt5$$
A: As was already mentioned, we have
$$0=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$$
$$|x|=|e^{2\pi i/5}|=1\implies x^2=x^{-3}\;,\;x=x^{-4}\;,\;x^{-k}=\overline{x^k}\implies$$
$$(x^2-x^3)^2(x^4-x)^2=(x^2-x^{-2})^2(x^{-1}-x)^2=(x-x^{-1})^4(x+x^{-1})^2=$$
$$=\left(2i\sin\frac{2\pi}5\right)^4\left(2\cos\frac{2\pi}5\right)^2=64\sin^4\frac{2\pi}5\cos^2\frac{2\pi}5=$$
$$=64\left(\frac14\sqrt{10+2\sqrt5}\right)^4\left(\frac14\left(-1+\sqrt5\right)\right)^2=5$$
A: $x^4+x^3+x^2+x+1=0$. 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 $z_2=2cos144^\circ=\dfrac{-1-\sqrt5}{2}$.
