Key Concepts and Formulas

• $A^B = 1$ conditions: $A^B = 1$ if $A=1$, $B=0$ (and $A \ne 0$), or $A=-1$ and $B$ is an even integer.
• Quadratic equation factoring: Factoring quadratic equations of the form $ax^2 + bx + c = 0$ to find roots.
• Integer parity: Understanding the difference between even and odd integers.

Step-by-Step Solution

Step 1: Identify the Base and Exponent

The given equation is $(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$. Here, the base $A = x^2 - 5x + 5$ and the exponent $B = x^2 + 4x - 60$. We need to find all real values of $x$ that satisfy this equation.

Step 2: Case 1: Base equals 1

• We set the base equal to 1: $x^2 - 5x + 5 = 1$ This is because $1$ raised to any power is $1$.
• Simplify the equation: $x^2 - 5x + 4 = 0$
• Factor the quadratic: $(x - 1)(x - 4) = 0$
• Solve for $x$: $x = 1 \quad \text{or} \quad x = 4$

Step 3: Case 2: Exponent equals 0

• We set the exponent equal to 0: $x^2 + 4x - 60 = 0$ This is because any non-zero number raised to the power of $0$ is $1$.
• Factor the quadratic: $(x - 6)(x + 10) = 0$
• Solve for $x$: $x = 6 \quad \text{or} \quad x = -10$
• Check if the base is non-zero for these values:
  • For $x = 6$: $A = (6)^2 - 5(6) + 5 = 36 - 30 + 5 = 11 \neq 0$.
  • For $x = -10$: $A = (-10)^2 - 5(-10) + 5 = 100 + 50 + 5 = 155 \neq 0$. Since the base is non-zero in both cases, both $x=6$ and $x=-10$ are valid solutions.

Step 4: Case 3: Base equals -1 and Exponent is an even integer

• We set the base equal to -1: $x^2 - 5x + 5 = -1$ This is because $(-1)^{\text{even integer}} = 1$.
• Simplify the equation: $x^2 - 5x + 6 = 0$
• Factor the quadratic: $(x - 2)(x - 3) = 0$
• Solve for $x$: $x = 2 \quad \text{or} \quad x = 3$
• Check if the exponent is an even integer for these values:
  • For $x = 2$: $B = (2)^2 + 4(2) - 60 = 4 + 8 - 60 = -48$, which is an even integer. So, $x=2$ is a valid solution.
  • For $x = 3$: $B = (3)^2 + 4(3) - 60 = 9 + 12 - 60 = -39$, which is an odd integer. So, $x=3$ is not a valid solution.

Step 5: Consolidate Solutions

The valid solutions are $x = 1, 4, 6, -10, 2$.

Step 6: Calculate the Sum

The sum of all real values of $x$ is $1 + 4 + 6 + (-10) + 2 = 3$.

Common Mistakes & Tips

• Forgetting the $0^0$ check: Always check if the base is non-zero when the exponent is zero.
• Parity check for base -1: Always verify that the exponent is an even integer when the base is -1.
• Missing cases: Remember to consider all three cases for $A^B = 1$.

Summary

To solve the equation $(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$, we considered three cases: the base equals 1, the exponent equals 0, and the base equals -1 with an even integer exponent. By solving the corresponding quadratic equations and checking for extraneous solutions (like $0^0$ or $(-1)^{\text{odd integer}}$), we found the valid solutions to be $x = 1, 4, 6, -10, 2$. The sum of these values is $3$.

Final Answer

The final answer is \boxed{3}, which corresponds to option (C).