How to solve $x^3 = x^{x^2-2x}$? Well i guess this is somehow pretty easy but there is something i don't understand.

I know that if $\;x&gt;0\;$ then I can compare the exponents: $\; 3=x^2-2x$ , and from here I get that $\;x=3\; or \;x=-1\;$, but because $\;x&gt;0\;$ that leaves me only with $\;x=3$ .

Second thing is that if both bases from both sides are equal to $1$ then its another solution, and therefore $\;x=1\;$ is another solution.

For conclusion we get that the solutions are: $\;x=3\; or \;x=1$ . 

Now my question is why $\;x=-1\;$ is also a solution? How do i get to this solution? am I suppose to just try place it and check because I somehow got it for $\;x&gt;0$ ? are there any steps I can follow for solving this kind of equations?

Another question is how do you call this kind of equation? when I was looking for "exponential equations" I could only find ones with numbers in the bases .

Thanks!
$x^a=x^b\implies a=b$ for all $x$,whether $x&gt;0$ or $x&lt;0$. You got $3=x^2-2x$, and solved this quadratic equation to get $x\in\{3,-1\}$. So $-1$ is indeed a solution. It depends on the equation as to which methods will you follow, there is no general algorithm. But yes, keep using exponent rules, and you will eventually lead to a solution.
look on rh(n)
and get x^2-2*x=3 (mod n)
Because rh(2)=-1 then we get 
(-1)^2+2=3=3 (mod 2)
and thats why it true