The number of real roots of $(6-x{)}^4+(8-x{)}^4=16$, is

Let $y=7-x$. Then, the given equation becomes

$\begin{array}{l}(y+1{)}^4+(y-1{)}^4=16\\ \Rightarrow y^4+6y^2-7=0\\ \Rightarrow \left(y^2-1\right)\left(y^2+7\right)=0\\ \Rightarrow y^2-1=0 \left[\because y^2+7\ne 0\right]\end{array}$

$\Rightarrow y=\pm 1\Rightarrow 7-x=\pm 1\Rightarrow x=6,8$