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?

You have
$$\begin{align}
y & = \frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} \\
& = \frac{3(1-3^{2-x}) + 3(1-3^{x-2})}{(1-3^{x-2})(1-3^{2-x})} \\
& = \frac{3(1-3^{2-x} + 1-3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{x-2+2-x})} \\
& = \frac{3(2-3^{2-x} - 3^{x-2})}{(1 - 3^{2-x}-3^{x-2}+3^{0})} \\
& = \frac{3(2-3^{2-x} - 3^{x-2})}{(2 - 3^{2-x}-3^{x-2})} \\
& = 3 & \text{since } x \not =2\\
\end{align}$$
So 
$$y^{y^2-6} = 3^{3^2-6}=3^{9-6}=3^3=27$$
And you're done.
