Giải phương trình: x³ ‒ 3x ‒ 4 = 0.

Đặt \[x = a + \frac{1}{a}\] (a ≠ 0)

Khi đó:

x³ ‒ 3x ‒ 4 = 0

\[ \Leftrightarrow {\left( {a + \frac{1}{a}} \right)^3} - 3\left( {a + \frac{1}{a}} \right) - 4 = 0\]

\[ \Leftrightarrow {a^3} + \frac{1}{{{a^3}}} - 4 = 0\]

⇔ a⁶ ‒ 4a³ + 1 = 0

⇔ (a³ ‒ 2)² = 3

\[ \Rightarrow \left[ \begin{array}{l}{a^3} = 2 + \sqrt 3 \\{a^3} = 2 - \sqrt 3 \end{array} \right.\]\[ \Rightarrow \left[ \begin{array}{l}a = \sqrt[3]{{2 + \sqrt 3 }}\\a = \sqrt[3]{{2 - \sqrt 3 }}\end{array} \right.\]

\[ \Rightarrow \left[ \begin{array}{l}x = \sqrt[3]{{2 + \sqrt 3 }} + \frac{1}{{\sqrt[3]{{2 + \sqrt 3 }}}}\\x = \sqrt[3]{{2 - \sqrt 3 }} + \frac{1}{{\sqrt[3]{{2 + \sqrt 3 }}}}\end{array} \right.\]