ĐK: $x\geq -2$
$\bigstar$ Với $x>2\rightarrow VT=x+x(x^2-4)>2x>\sqrt{x+2}\rightarrow $ PT vô nghiệm.
$\bigstar$ Với $x \in [-2;2],$ set $x=2cost$ $(t\in[0;\pi])$
$\rightarrow \sqrt{x+2}=\sqrt{2(1+cost)}=\sqrt{2.2cos^2\frac{t}{2}}=2|cos\frac{t}{2}|=2cos\frac{t}{2}$ ( do $t \in [0;\frac{\pi}{4}])$
PT $\Leftrightarrow 2(4cos^3t-3cost)=2cos\frac{t}{2}$
$\Leftrightarrow 8cos^3t-6cost=2cos\frac{t}{2}$
$\Leftrightarrow cos3t=cos\frac{t}{2}$
$\Leftrightarrow 3t=\frac{t}{2}+k2\pi$ v $3t=\frac{-t}{2}+k2\pi$ $(k \in Z)$
$\Leftrightarrow t=\frac{k4\pi}{5}$ v $t=\frac{k4\pi}{7}$
Do $\left\{ \begin{array}{l} t \in [0;\pi]\\ k \in Z \end{array} \right.\Rightarrow t \in {0;\frac{4 \pi}{7};\frac{4 \pi }{5}}$
$\Rightarrow x={2;2cos\frac{4 \pi }{7};2cos\frac{4 \pi }{5}}$