$\ (x - 1)^{x + 2} = (x - 1)^{x + 6}$
$\ ⇒ (x - 1)^{x + 6} - (x - 1)^{x + 2} = 0$
$\ ⇒ (x - 1)^{x + 2} . (x -1)^{x + 4} - (x - 1)^{x + 2} = 0$
$\ ⇒ (x - 1)^{x + 2}. [(x - 1)^{4} - 1] = 0$
$\ ⇒ \left[ \begin{array}{l}(x - 1)^{x + 2} = 0\\(x - 1)^{4} - 1 = 0\end{array} \right.$
$\ ⇒ \left[ \begin{array}{l}x = 1\\\text{x ∈ { 2 ; 0 }}\end{array} \right.$
Vậy $\text{x ∈ { 1 ; 2 ; 0 }}$