`(x-6)^3` = `(x-6)^2`
`=>` `(x-6)^3` - `(x-6)^2` = 0
`=>` `(x-6)^2` . [ ( x - 6 ) - 1 ] = 0
`=>` `(x-6)^2` . ( x - 7 ) = 0
`=>` \(\left[ \begin{array}{l}(x-6)^2=0\\x-7=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x-6=0\\x-7=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=6\\x=7\end{array} \right.\)
Vậy x ∈ { 6; 7 }