How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?

Calling $z = 3^{x-2}$ we have
$$
3\left(\frac{1}{1-z}+\frac{1}{1-z^{-1}}\right) = y \Rightarrow y = 3
$$
so finally
$$y^{y^2-6} = 3^{9-6} = 27$$
