I'm feeling a little silly askign this question, but after about 2 hours of circling around the same point I am getting frustrated.
Starting with the expressions for $M$ and $R$ from the lecture notes: \begin{align*} M=4\pi\left[\frac{K(n+1)}{4\pi{G}}\right]^{3/2}\rho_{c}^{(3-n)/2n}\left[-\xi^{2}\frac{d\theta}{d\xi}\right] \end{align*} \begin{align*} R=\left[\frac{K(n+1)}{4\pi{G}}\right]^{1/2}\rho_{c}^{(1-n)/2n} \end{align*} I'm assuming the aim of the expression wanted in the question is to remove the dependency of both $R$ and $M$ on the core density, $\rho_{c}$, to do that just requires a comparison of the power terms for both $\rho_{c}$ variables in each expression. Therefore: \begin{align*} M^{(n-1)}=&(4\pi)^{(n-1)}\left[\frac{K(n+1)}{4\pi{G}}\right]^{3(n-1)/2}\rho_{c}^{(n-1)(3-n)/2n}\left[-\xi^{2}\frac{d\theta}{d\xi}\right]^{(n-1)}\\\\ =&\left[4\pi\left(-\xi^{2}\frac{d\theta}{d\xi}\right)\right]^{(n-1)}\left[\frac{K(n+1)}{4\pi{G}}\right]^{3(n-1)/2}\rho_{c}^{(n-1)(3-n)/2n} \end{align*} Same treatment for $R$: \begin{align*} R^{(3-n)}=\left[\frac{K(n+1)}{4\pi{G}}\right]^{(3-n)/2}\rho_{c}^{(1-n)(3-n)/2n} \end{align*}
Essentially, for some strange reason I am having troubles eliminating $\rho_c$. I want to get:
$$ M^{(n-1)}R^{(3-n)} $$
This should cancel out the $\rho_c$ term and also give the bracket with K within it to the power $n$. Could anyone run though this and let me know if it is possible to end up with $\rho_{c}^{0}$ (i.e., cancelling it out), or do I need to introduce something else to make it consistent? Very confused at how this is achieved.
You have $$M^{n-1}=A\rho_{c}^{(n-1)(3-n)/2n}\quad\Rightarrow\quad \frac{M^{n-1}}{A}=\rho_{c}^{(n-1)(3-n)/2n}$$ where $$A=\left[4\pi\left(-\xi^{2}\frac{d\theta}{d\xi}\right)\right]^{(n-1)}\left[\frac{K(n+1)}{4\pi{G}}\right]^{3(n-1)/2}$$ and $$R^{3-n}=B\rho_{c}^{(1-n)(3-n)/2n}\quad\Rightarrow\quad \frac{B}{R^{3-n}}=\rho_{c}^{(n-1)(3-n)/2n}$$ where $$B=\left[\frac{K(n+1)}{4\pi{G}}\right]^{(3-n)/2}.$$
Hence, you can have $$\frac{M^{n-1}}{A}=\frac{B}{R^{3-n}}.$$