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

Solution :
$${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$$
Case I
$${x^2} - 5x + 5 = 1\,\,{\text{and }}{x^2} + 4x - 60$$       can be any real number
$$ \Rightarrow \,\,x = 1,4$$
Case II
$${x^2} - 5x + 5 = - 1\,\,{\text{and }}{x^2} + 4x - 60$$       has to be an even number
$$ \Rightarrow \,\,x = 2,3$$
where 3 is rejected because for $$x = 3,{x^2} + 4x - 60$$    is odd.
Case III
$${x^2} - 5x + 5$$   can be any real number and $${x^2} + 4x - 60 = 0$$
$$ \Rightarrow \,\,x = - 10,6$$
⇒ Sum of all values of $$x = - 10 + 6 + 2 + 1 + 4 = 3$$