[tex]\bf \begin{cases}
(f\circ g)(x)=x^2-6x+8\\\\
(f\circ g)(x)=f(~~g(x)~~)\\\\
g(x)=x-3\\
\end{cases}
\\\\\\
\textit{now, one probability is }x^2-1
\\\\\\
f(x)=x^2-1\implies f(~~g(x)~~)=(x-3)^2-1
\\\\\\
f(~~g(x)~~)=(x^2-6x+9)-1\implies f(~~g(x)~~)=x^2-6x+8[/tex]