Reading a paper on stochastic dynamic optimisation, I got stuck on a (seemingly) simple algebraic step. The starting equation is as follows:
$$\rho (\alpha+\theta N^{1-\eta})=\frac{1}{1-\eta}\left((\theta(1-\eta))^{\frac{\eta-1}{\eta}}N^{1-\eta}-1\right) \\ +\left(\theta(1-\eta)N^{1-\eta}\right)\left(r-(\theta(1-\eta))^{-\frac{1}{\eta}}-\frac{1}{2\eta}\left(\frac{\mu-r}{\sigma}\right)^{2}\right) $$
Then, by setting $\alpha\rho(1-\eta)+1=0$ the author eliminates $^{1-\eta}$ and obtains an equation in $\theta$, which, when solved, has the following result:
$$\theta^{*}=\frac{1}{1-\eta}\left[\frac{\rho+r(1-\eta)}{\eta}+\frac{(1-\eta)}{2\eta^{2}}\left(\frac{\mu-r}{\sigma}\right)^{2}\right]^{-\eta} $$
It is probably something very simple but I cannot figure out how to eliminate $^{1-\eta}$ using $\alpha\rho(1-\eta)+1=0$.
When you look closely, you will see that the equation is of the form $\alpha\rho +N^{1-\eta}(\mathrm{something})=-\frac{1}{1-\eta}+N^{1-\eta}(\mathrm{something})$, so when $\alpha\rho+\frac{1}{1-\eta}=0$, this equation is of the form $N^{1-\eta}(\mathrm{something})=0$ which has solution $\mathrm{something}=0$, which has eliminated $N^{1-\eta}$.
Explicitly, dropping the $\rho\alpha$ on the LHS and $-\frac1{1-\eta}$ on the RHS we obtain:
$$\rho \theta N^{1-\eta}=\frac{1}{1-\eta}(\theta(1-\eta))^{\frac{\eta-1}{\eta}}N^{1-\eta} +\left(\theta(1-\eta)N^{1-\eta}\right)\left(r-(\theta(1-\eta))^{-\frac{1}{\eta}}-\frac{1}{2\eta}\left(\frac{\mu-r}{\sigma}\right)^{2}\right) $$ Note that the LHS is just $N^{1-\eta}\rho\theta$ and the RHS is equal to $$N^{1-\eta}\left(\frac{1}{1-\eta}(\theta(1-\eta))^{\frac{\eta-1}{\eta}} +\theta(1-\eta)\left(r-(\theta(1-\eta))^{-\frac{1}{\eta}}-\frac{1}{2\eta}\left(\frac{\mu-r}{\sigma}\right)^{2}\right)\right)$$
So dividing LHS and RHS by $N^{1-\eta}$ gives
$$\rho\theta=\frac{1}{1-\eta}(\theta(1-\eta))^{\frac{\eta-1}{\eta}} +\theta(1-\eta)\left(r-(\theta(1-\eta))^{-\frac{1}{\eta}}-\frac{1}{2\eta}\left(\frac{\mu-r}{\sigma}\right)^{2}\right) $$
We can then indeed simplify to $$\theta\rho=\frac\eta{1-\eta}\big(\theta(1-\eta)\big)^{1-\frac1\eta}+\theta(1-\eta)\left(r-\frac1{2\eta}\left(\frac{\mu-r}\sigma\right)^2\right)$$
Dividing by $\theta$ (Or alternatively, noting that $\theta=0$ is also a solution), we get the equation $$\rho=\eta\big(\theta(1-\eta)\big)^{-\frac1\eta}+(1-\eta)\left(r-\frac1{2\eta}\left(\frac{\mu-r}\sigma\right)^2\right)$$ for the non-trivial solutions for $\theta$. You can solve this in the standard manner: Move $(1-\eta)\left(r-\frac1{2\eta}\left(\frac{\mu-r}\sigma\right)^2\right)$ to the left, divide both sides by $\eta$, take both sides to the power of $-\eta$, then divide by $1-\eta$ and you indeed obtain $$\frac{1}{1-\eta}\left[\frac1\eta\left(\rho-(1-\eta)\left(r-\frac1{2\eta}\left(\frac{\mu-r}\sigma\right)^2\right)\right)\right]^{-\eta}=\theta$$
which is the desired equation
(the one in the pdf you linked in the comment. The equation in the question has $\frac{\rho+r(1-\eta)}\eta$ instead of $\frac{\rho+r(\eta-1)}\eta)$.
This is not a homework question I am doing for you right?