Given $f(x)=(x-5)^{4}$
Differentiate with respect to $x$
$
f(x)=4(x-5)^{2}
$
For local maxima and minima
$
\begin{array}{l}
f(x)=0 \\
=4(x-5)^{2}=0 \\
=x-5=0 \\
x=5
\end{array}
$
$f(x)$ changes from negative to positive as passes through $5 .$
So, $x=5$ is the point of local minima
Thus, local minima value is $f(5)=0$