The sum of all real values of $$x$$ satisfying the equation $$\left(x^2 - 5x + 5\right)^{x^2 + 4x - 60} = 1$$ is

We begin with the equation

$$\left(x^{2}-5x+5\right)^{\,x^{2}+4x-60}=1.$$

For any real numbers $$a$$ and $$b$$, the expression $$a^{b}$$ equals $$1$$ in the following real-number situations:

$$\begin{aligned} \text{(i)}\;& b=0 &\text{and}&\; a\neq0,\; &\text{because}&\; a^{0}=1,\\[2pt] \text{(ii)}\;& a=1 &\text{with}&\; b\in\mathbb R,\; &\text{because}&\; 1^{b}=1,\\[2pt] \text{(iii)}\;& a=-1 &\text{with}&\; b=\text{even integer},\; &\text{because}&\;(-1)^{\text{even}}=1. \end{aligned}$$

We shall apply each case to the base

$$f(x)=x^{2}-5x+5$$

and the exponent

$$g(x)=x^{2}+4x-60.$$

Case 1: Exponent zero. We set $$g(x)=0$$ and require $$f(x)\neq0.$$

$$x^{2}+4x-60=0.$$

The discriminant is $$\Delta=4^{2}-4(1)(-60)=16+240=256,$$ so $$\sqrt{\Delta}=16.$$

$$x=\frac{-4\pm16}{2}\;\Rightarrow\;x=\frac{12}{2}=6,\quad x=\frac{-20}{2}=-10.$$

Checking the base:

$$f(6)=36-30+5=11\neq0,$$, $$f(-10)=100+50+5=155\neq0.$$

Thus $$x=6$$ and $$x=-10$$ satisfy the equation.

Case 2: Base equal to 1. We set $$f(x)=1$$ (no restriction on $$g(x)$$).

$$x^{2}-5x+5=1\;\Rightarrow\;x^{2}-5x+4=0.$$

The discriminant is $$25-16=9,$$ so $$\sqrt{\Delta}=3.$$

$$x=\frac{5\pm3}{2}\;\Rightarrow\;x=4,\quad x=1.$$

Hence $$x=1$$ and $$x=4$$ are solutions.

Case 3: Base equal to $$-1$$ and exponent an even integer.

First make the base $$-1$$:

$$x^{2}-5x+5=-1\;\Rightarrow\;x^{2}-5x+6=0.$$

The discriminant is $$25-24=1,$$ so $$\sqrt{\Delta}=1.$$

$$x=\frac{5\pm1}{2}\;\Rightarrow\;x=3,\quad x=2.$$

Now we need $$g(x)$$ to be an even integer at these $$x$$ values.

For $$x=3:$$ $$g(3)=3^{2}+4\cdot3-60=9+12-60=-39,$$ which is odd, so $$x=3$$ is rejected.

For $$x=2:$$ $$g(2)=2^{2}+4\cdot2-60=4+8-60=-48,$$ which is an even integer, so $$x=2$$ is accepted.

No other cases produce real solutions equal to $$1,$$ so the complete set of real solutions is

$$\{\, -10,\; 6,\; 1,\; 4,\; 2 \,\}.$$

The problem asks for the sum of all these real values:

$$\begin{aligned} -10+6+1+4+2 &= (-10+6)+1+4+2\\ &= -4+1+4+2\\ &= (-4+1)+4+2\\ &= -3+4+2\\ &= 1+2\\ &= 3. \end{aligned}$$

Hence, the correct answer is Option C.