Polynomial Power Polynomial?

Let $A = x^3 - 12x^2 - x + 133$ and $B = x^2 -6x - 55$ , then we have the equation $A^B = 1$ . So all of the solutions to this equations must satisfy at least one of the following cases:

Case one : $A = 1$ .
Case two : $B = 0$ but $A \ne 0$ .
Case three : $A = -1$ , and $B$ is an even number.

For case one :, we have

$x^3 - 12x^2 - x + 133 = 1\Rightarrow x^3-12x^2 - x + 132 = 0$

By rational root theorem , a possible root of the cubic equation above is $\pm \; \text{factors of 132}$ . Trial and error shows that $x = -3$ is a possible solution. By performing long division of polynomials, we have

$x^3-12x^2 - x + 132 = (x+3)(x^2 - 15x + 44) = (x+3)(x-11)(x-4) = 0 .$

Thus the roots are $x=-3,11,4$ . Substituting each of these values individually shows that all these 3 values are possible solutions.

For case two :, we have $x^2 -6x - 55 = 0 \Rightarrow (x+5)(x-11) = 0 \Rightarrow x = -5, 11$ .

Substituting each of these values individually shows that these 2 values are possible solutions as well.

For case three , we have $x^3 - 12x^2 - x + 133 = -1 \Rightarrow x^3 - 12x^2 - x + 134 = 0$ . Rational root theorem shows that there is no integer solution of $x$ , thus it is impossible for any of 3 roots of this cubic equation to satisfy the equation $x^2 - 6x - 55 = \text{ some even number}$ .

In other words, there is no solution for case three.

Hence, gathering all the solutions from cases one to three, we have $x = -3,-5,4,11$ , and their sum is $-3-5+4+11 = \boxed7$ .

The $LHS = 1$ when:

$\begin{cases} x^3 - 12x^2 -x+133 = 1 & ...(1) \\ x^2-6x-55 = 0 & ...(2) \end{cases}$

$\begin{aligned} (1): \quad x^3 - 12x^2 -x+132 & = 0 \quad \quad \small \color{#3D99F6}{\text{By rational root theorem, we find that 11 is a root}} \\ (x-11)(x^2 - x -12) & = 0 \\ (x-11)(x+3)(x-4) & = 0 \\ \Rightarrow x & = \begin{cases} -3 \\ 4 \\ 11 \end{cases} \end{aligned}$

$\begin{aligned} (2): \quad x^2-6x-55 & = 0 \\ (x+5)(x-11) & = 0 \\ \Rightarrow x & = \begin{cases} -5 \\ 11 \end{cases} \end{aligned}$

Therefore, there are $4$ distinct solutions and their sum $= -5-3+4+11 = \boxed{7}$