Đáp án:
$\begin{array}{l}
{x^3} + {x^2} + x = - \dfrac{1}{3}\\
\Rightarrow 3{x^3} + 3{x^2} + 3x + 1 = 0\\
\Rightarrow 2{x^3} + {x^3} + 3{x^2} + 3x + 1 = 0\\
\Rightarrow 2{x^3} + {\left( {x + 1} \right)^3} = 0\\
\Rightarrow {\left( {\sqrt[3]{2}x} \right)^3} + {\left( {x + 1} \right)^3} = 0\\
\Rightarrow \left( {\sqrt[3]{2}x + x + 1} \right)\left( {\sqrt[3]{4}{x^2} - \sqrt[3]{2}.x.\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right) = 0\\
Do:\left( {\sqrt[3]{4}{x^2} - \sqrt[3]{2}.x.\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2} = 0} \right)\text{vô nghiệm}\\
\Rightarrow \sqrt[3]{2}.x + x + 1 = 0\\
\Rightarrow \left( {\sqrt[3]{2} + 1} \right).x = - 1\\
\Rightarrow x = \dfrac{{ - 1}}{{\sqrt[3]{2} + 1}} = \dfrac{{ - 1.\left( {\sqrt[3]{4} - \sqrt[3]{2} + 1} \right)}}{{{{\left( {\sqrt[3]{2}} \right)}^3} - 1}}\\
= \dfrac{{ - \sqrt[3]{4} + \sqrt[3]{2} - 1}}{1}\\
= - \sqrt[3]{4} + \sqrt[3]{2} - 1\\
\text{Vậy}\,x = - \sqrt[3]{4} + \sqrt[3]{2} - 1
\end{array}$
