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

Case 1 : When x² - 5x + 5 = 1 and x² + 4x - 60 is any real no then this equation satisfy.
Note : When we put any real number as a power of 1 the value stays always 1 (1 ^{any real no} = 1).
x² - 5x + 5 = 1
(x - 1)(x - 4) = 0
$$\therefore$$ x = 1, 4

Case 2 : When x² - 5x + 5 is a real no and x² + 4x - 60 = 0 then the given equation satisfy. As we know if power of any real no is zero then it will become 1((any real number)⁰ = 1).
For, x² + 4x - 60 = 0
(x - 6)(x + 10) = 0
$$\therefore$$ x = 6, -10

Case 3 : When x² - 5x + 5 = -1 and x² + 4x - 60 is even this equation satisfy. As we know (-1)^even = 1.
For, x² - 5x + 5 = -1
(x - 2)(x - 3) = 0
$$\therefore$$ x = 2, 3
But x can't be 3 because when x = 3 the value of x² + 4x - 60 becomes 3² + 4.3 - 60 = - 39 which is an odd number, then (-1)^-39 = -1. So for x = 3 equation does not satisfy.

$$\therefore$$ The sum of all the real values = 1 + 4 + 6 + (-10) + 2 = 3