Giải phương trình: x^3 – 7x – 6 = 0

x³ – 7x – 6 = 0

⇔ x³ – 4x – 3x – 6 = 0

⇔ x(x² – 4) – 3(x + 2) = 0

⇔ x(x – 2)(x + 2) – 3(x + 2) = 0

⇔ (x + 2)(x² – 2x – 3) = 0

⇔ (x + 2)(x + 1)(x – 3) = 0

⇔ \(\left[ \begin{array}{l}x = - 2\\x = - 1\\x = 3\end{array} \right.\)

Vậy x ∈ {-2;-1;3}.