if w is a non real cube root of unity prove that (2+5w+2w^2)^6 =729
i am not really sure how to do this
$w$ is a cube root of unity so we know it's magnitude is $1$
Thus we can write $w = e^{i \theta}$
let $f(w) = 2 + 5w + 2w^2$
a little work shows that
$f(e^{i\theta}) = (5+4\cos(\theta))e^{i \theta}$
and thus
$\left(f(w)\right)^6 = (5+4\cos(\theta))^6 e^{6i \theta}= 729$
and as $ (5+4\cos(\theta))^6 \in \mathbb{R}$ it must be that
$(5+4\cos(\theta))^6 = 729$
$5+4\cos(\theta) = 3$
$\cos(\theta) = -\dfrac 1 2$
$\theta = \dfrac{2\pi}{3},~\dfrac{4\pi}{3}$
$w = e^{i2\pi/3},~e^{i 4\pi/3}$
and these are exactly the two non-real cube roots of unity.
Actually you'll need to reverse these steps as you are given $w$ is a cube root of unity and must show that $f(w)^6 = 729$
It pretty much just amounts to plugging in the two non-real cube roots of unity and grinding through the algebra using the one shortcut I noted.
There are two non-real cube roots of unity: $\omega=-\frac{1}{2}+\bf{i}\frac{\sqrt3}{2}$ is one of them.
Now $\omega^2=-\frac{1}{2}-\bf{i}\frac{\sqrt3}{2}$ Can you show that?
You calculate $2+5\omega +2\omega^2$. If that is $3$ you are done because $3^6=729$.
$\displaystyle \omega ^3-1=(\omega -1)\left(\omega ^2+\omega +1\right)$
and since $\displaystyle \omega \neq 1$ we get $\displaystyle \omega ^2+\omega +1=0$
so that $\displaystyle 2\omega ^2+2\omega +2=0$ and $\displaystyle 2+5\omega +2\omega ^2=3\omega $
$\displaystyle \left(2+5\omega +2\omega ^2\right)^6=3^6\omega ^6=729$
That was given.
There are exactly three cube roots of unity, two of which are non-real:
$\large{\exp \left( {\dfrac{{2\pi\bf{i} }}{{\sqrt 3 }}} \right) = \cos \left( {\dfrac{{2\pi }}{{\sqrt 3 }}} \right) \pm\bf{i} \sin \left( {\dfrac{{2\pi }}{{\sqrt 3 }}} \right) = \dfrac{1}{2} \pm \bf{i}\dfrac{{\sqrt 3 }}{2}}$
Those are the only two non-real.