Tìm x, biết: x³ – 16x = 0.

Ta có:

x³ – 16x = 0

⇔ x(x² – 16) = 0

⇔ x(x – 4)(x + 4) = 0

\( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x + 4 = 0\\x - 4 = 0\end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = - 4\\x = 4\end{array} \right.\)

Vậy x ∈ {0; 4; –4}.