Rút gọn biểu thức $A = \sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}$ ta được:

Ta có: $A = \sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}$

Suy ra: ${A^3} = {\left( {\sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}} \right)^3}$$= {\left( {\sqrt[3]{{9 + 4\sqrt 5 }}} \right)^3} + {\left( {\sqrt[3]{{9 - 4\sqrt 5 }}} \right)^3} + 3\sqrt[3]{{9 + 4\sqrt 5 }}.\sqrt[3]{{9 - 4\sqrt 5 }}\left( {\sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}} \right)$

$= 9 + 4\sqrt 5  + 9 - 4\sqrt 5  + 3.\sqrt[3]{{\left( {9 + 4\sqrt 5 } \right)\left( {9 - 4\sqrt 5 } \right)}}.A$  (vì $A = \sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}$)

$= 18 + 3\sqrt[3]{{{9^2} - {{\left( {4\sqrt 5 } \right)}^2}}}.A$$= 18 + 3A$

Hay ${A^3} = 3A + 18 \Leftrightarrow {A^3} - 3A - 18 = 0$$\Leftrightarrow {A^3} - 27 - 3A + 9 = 0$

$\Leftrightarrow \left( {A - 3} \right)\left( {{A^2} + 3A + 9} \right) - 3\left( {A - 3} \right) = 0$$\Leftrightarrow \left( {A - 3} \right)\left( {{A^2} + 3A + 6} \right) = 0$

$\Leftrightarrow \left[ \begin{array}{l}A - 3 = 0\\{A^2} + 3A + 6 = 0\end{array} \right. \\\Leftrightarrow \left[ \begin{array}{l}A = 3\\{\left( {A + \dfrac{3}{2}} \right)^2} + \dfrac{{15}}{4} = 0\left( {VN} \right)\end{array} \right.$

Vậy $A = 3.$