Ugly expression. Cant tell if I can simplify. Is there a general method for simplifying basic algebraic expressions?

Question

Let $\rho, \omega, c > 0$ and let $\alpha \in [0,1]$. I have managed to calculate

\begin{align*} \frac{\rho}{\alpha x^{\alpha-1}y^{1-\alpha}}&= \eta\\ \frac{\omega}{(1-\alpha) x^{\alpha}y^{-\alpha}}&= \eta\\ \left(\frac{\rho}{\omega} \frac{1-\alpha}{\alpha}\right)^{\alpha-1}c &= x \\ \left(\frac{\rho}{\omega} \frac{1-\alpha}{\alpha}\right)^{\alpha}c &=y\\ \end{align*}

So I obviously tried plugging in. But this lead me to a really ugly expression:

\begin{align*} \eta&=\frac{\rho}{\alpha \left(\left(\frac{\rho}{\omega} \frac{1-\alpha}{\alpha}\right)^{\alpha-1}c\right) ^{\alpha-1} \left(\left(\frac{\rho}{\omega} \frac{1-\alpha}{\alpha}\right)^{\alpha}c\right) ^{1-\alpha}}\\ &= \frac{\rho}{\alpha \left(\frac{\rho}{\omega} \frac{1-\alpha}{\alpha}\right)^{(2\alpha-1)(\alpha-1)} }\\ \end{align*} My Question:

(1) How can I simplify this?

(2) Is there a general method for how to simplify an arithmetic expression? I seem to have a lot of these formulas that are quite complicated and it would be nice if there were some other way besides trial and error.

Answer 1

I think you made a mistake in your calculation. The correct answer will be $$\eta=\frac{\rho}{\alpha \left(\frac{\rho}{\omega} \frac{1-\alpha}{\alpha}\right)^{\color{red}{1-\alpha}}}. $$ I don't know if you think this is simple.

Answer 2

Let $\frac{\rho}{\omega}(1/\alpha-1)=a$, then, you have $x=a^{\alpha-1}c,\ y=a^\alpha c$. So, $$\eta=\frac{\rho}{\alpha a^{1-\alpha}}=\frac{\omega a^\alpha}{1-\alpha}$$

Answer 3

Try looking at $x/y$. I note that you first two equations involve $x/y$. Set $z=x/y$ and then use the third and fourth equations to deduce

$$z = \frac{\omega}{\rho}\frac{\alpha}{1-\alpha}$$

Equation 1 and 2 become

$$\eta = \frac{\omega}{(1-\alpha) z^{\alpha}}$$

And

$$\eta = \frac{\rho}{\alpha z^{(\alpha-1)}}$$

So

$$\eta = \frac{\omega}{(1-\alpha)(\frac{\omega\alpha}{\rho(1-\alpha)})^{\alpha}}$$

And therefore

$$\eta = \frac{\rho^\alpha(1-\alpha)^{(\alpha-1)}}{(\omega^{(\alpha-1)}\alpha^\alpha)}$$