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

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

Case I: x²–5x+5=1 and x²+4x–60 can be any real number 

$\Rightarrow \mathrm{x}=1,4$

Case II: x²–5x+5=–1 and x²+4x–60 has to be an even number

$\Rightarrow \mathrm{x}=2,3$ (Where 3 is rejected because x²+4x–60 is odd)

Case III: x²–5x+5=1 can be any real number and x²+4x–60 = 0

$\Rightarrow \mathrm{x}=-10,6\Rightarrow \text{ Sum of the values of }$

$\mathrm{x}=-10+6+2+1+4=3$