Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$ I have this equation from this paper (Q.63)

Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.

My attempt-
$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$
So, $x=6$ and $x=-10$ makes the power $0$ and hence the equation becomes one. So, the sum of the real values of $x$ satisfying the equation is $6+(-10)=-4$. But answer is $+3$.
Where have I gone wrong?

You missed some cases.
The cases are: $\text{number}^0,1^\text{number},{-1}^{\text{even}}$
You have only considered the first case. 


*

*For first case, $$x^2+4x-60=0$$ the roots are $6,-10$.

*For the second case, $$x^2-5x+5=1$$ the roots are $1,4$.

*For the third case, $$x^2-5x+5=-1$$ $x$ can be $2$ or $3$. If you substitute the value $2$ on the equation $x^2+4x-60$, you get $-48$ which satisfies the condition since $-1^{-48}=1$. If you substitute $x=3$, you get $-39$ which does not satisfy the given condition since $-1^{-39}\neq 1$. So only $x=2$ satisfies the condition.
The real values of $x$ are therefore: $1,2,4,6,-10$. The sum of real values of $x$ is therefore $3$.
