Find the product of all the solutions of $\displaystyle\left(\frac{x^2-5x}{6}\right)^{x^2-2}=1$ times the number of solutions.
I don't know how to solve an exponential equation, so I've done as follow:
If you raise something to the $0$th power you get $1$, so:
$\begin{align*} &x^2 - 2 = 0\\ &(x+\sqrt{2})(x-\sqrt{2}) = 0\\ &x = \pm \sqrt{2} \end{align*}$If the result is $1$ then $\displaystyle\frac{x^2-5x}{6}=\pm1$. When it is equal to $1$ the exponent can be anything, if it is $-1$ it must be even. So:
$x^2-5x-6=0 \Rightarrow x_1 = -1, x_2 = 6$
$x^2 - 5x + 6 = 0$, $x_1 = 2 \Rightarrow x_2 = 3$ but $x=3$ is not acceptable because $x^2-2 = 7$, odd.
So the solutions are: $S=\{-\sqrt{2}, -1, 2, \sqrt{2}, 6\}$, and the answer to the problem $120$.
Is my work correct? Are there any other methods (simpler, complicated ones)?
EDIT: Wolfram|Alpha does not agree with me:
Wolfram|Alpha results