($x^{3}$ + $8$).($-2x-5$)=$0$
$⇒$ \(\left[ \begin{array}{l}x^{3} + 8 = 0\\-2x-5=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x^{3} = -8\\-2x=5\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x= -2\\x=\frac{5}{-2}\end{array} \right.\)
Vậy $x$ ∈ {$\frac{5}{-2}$;$-2$}