How do we express nonics in terms of the monic?

I'll be repeatedly using the fact that if $a+b+c=0$ ,then $a^3+b^3+c^3=3abc$ .

Here we go: $\begin{aligned} r+\frac{1}{r}+(-3)&=0\\ \implies r^3+\frac{1}{r^3}-27&=-9\\ r^3+\frac{1}{r^3}+(-18)&=0\\ \implies r^9+\frac{1}{r^9}-5832&=-54\\ \implies r^9+\frac{1}{r^9}&=\boxed{5778} \end{aligned}$

$r^3+ \dfrac 1 {r^3}=\left(r+\dfrac 1 r \right)^3-\left(3 \times r \times \dfrac 1 r \times \left(r+ \dfrac 1 r \right)\right)=3^3-3(3)=18$ .

Similarly, $r^9+\dfrac 1 {r^9}=\left(r^3+\dfrac 1 {r^3}\right)^3-\left(3 \times r^3 \times \dfrac 1 {r^3} \times \left(r^3+\dfrac 1 {r^3}\right)\right)=18^3-3(18)=\boxed{5778}$ .

Here's another way, let $r=e^{i\theta}$ where $\theta$ is a complex number.

Note that $e^{ix}=\cos x+ i\sin x \quad \forall x \in \mathbb{C}$

Then, we're given that $2\cos \theta = 3$ and we need to find $2\cos 9\theta$ . Using triple angle formulae (which is valid for complex arguements) twice, we get $2 \cos 9\theta =\boxed{5778}$

$\begin{aligned} r + \frac{1}{r} & = 3 \\ \left(r + \frac{1}{r}\right)^3 & = 3^3 \\ r^3+3r+\frac{3}{r}+\frac{1}{r^3} & = 27 \\ \implies r^3 + \frac{1}{r^3} & = 27 - 3 \left(r + \frac{1}{r}\right) \\ & = 27 - 9 = 18 \end{aligned}$

$\begin{aligned} \left(r^3 + \frac{1}{r^3}\right)^3 & = 18^3 \\ r^9+3r^3+\frac{3}{r^3}+\frac{1}{r^9} & = 5832 \\ \implies r^9 + \frac{1}{r^9} & = 5832 - 3 \left(r^3 + \frac{1}{r^3}\right) \\ & = 5832 - 54 = \boxed{5778} \end{aligned}$