What is sum of all solutions to equation: x^(x^2 + 2)

Key concept: If \(b^x = b^y\), then \(x = y \) (as long as \(b \neq 0\), \(b \neq 1\), and \(b \neq -1\))

The provisos here a very important.
For example, if \(0^x = 0^y\), we can't then conclude that \(x=y\)

Now on to the solution.....
If \(x^{(x^2 + 2)} = x^{(4x + 14)}\), then we can conclude that: \(x^2 + 2 = 4x + 14\)
Rewrite as follows: \(x^2 - 4x - 12= 0\)
Factor: \((x-6)(x+2)=0\)
So, two solutions are \(x=6 \) and \(x = -2\)

At this point, we need to test the "proviso" values (\(x=0\), \(x=1\) and \(x=-1\))

Test \(x=0\) to get: \(0^{0^2 + 2} = 0^{4(0) + 14}\). Works! So, another solution is \(x=0\)
Test \(x=1\) to get: \(1^{1^2 + 2} = 1^{4(1) + 14}\). Works! So, another solution is \(x=1\)
Test \(x=-1\) to get: \((-1)^{(-1)^2 + 2} = (-1)^{4(-1) + 14}\). DOESN'T work, So, \(x=-1\) is NOT a solution

So, the sum of the solutions \(= 6 + (-2) + 0 + 1 = 5\)

Answer: C