Find the sum of the real solutions to the equation $$(x^{2} - x - 1)^{x^{2} - x - 6} = 1$$

Question

anonymous · Answer

if the exponent is zero, then wouldn't it give you a true statement?
so solve $\large x^2-x-6=0 $

anonymous · Answer

of course, the base cannot be zero...

anonymous · Answer

Actually, it can because I believe 0^0 is still 1? I believe there are still 2 ways that the solution can be 1. One of which is if the base is 1. I'm trying to figure out the last one.

anonymous · Answer

in calculus that is known as indeterminate... 0^0

anonymous · Answer

Alright. It's still a situation where it would work though as long it's not 0, so it's a solution.

anonymous · Answer

oh yeah... i didn't think about tat one... letting the base = 1....

anonymous · Answer

Wait, isn't -1^(even power) possible?

anonymous · Answer

yes... you're correct.... 
wow... i gotta review my algebra!!!

anonymous · Answer

so to answer this question, are you gonna have to sum up the solutions from:
base = 1;
exponent = 0;
base = (-1) with an even power.???

anonymous · Answer

Alright. I got 2, -1, 3, -2, 0, 1 for the values of x. Then I just add them?

anonymous · Answer

that's what i think it says...

anonymous · Answer

Alright. Thanks!