\(\left(x-6\right)^3=\left(x-6\right)^2\)
Đặt \(x-6=a\)\(\Leftrightarrow a^3=a^2\)\(\Leftrightarrow a^3-a^2=0\)\(\Leftrightarrow a^2\left(a-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a^2=0\\a-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}a=0\\a=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-6=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=7\end{cases}}\)(nhận)
Vậy \(x\in\left\{6;7\right\}\)