Thực hiện phép tính

    a). \[A = \left( {\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}} \right)\left( {\sqrt[3]{3} - \sqrt[3]{2}} \right)\]                                            b). \[B = \sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}.\]

a). Ta có:\[A = \left( {\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}} \right)\left( {\sqrt[3]{3} - \sqrt[3]{2}} \right) = \left( {{{\left( {\sqrt[3]{3}} \right)}^2} + \sqrt[3]{3}\sqrt[3]{2} + {{\left( {\sqrt[3]{2}} \right)}^2}} \right)\left( {\sqrt[3]{3} - \sqrt[3]{2}} \right)\]

\[ = {\left( {\sqrt[3]{3}} \right)^3} - {\left( {\sqrt[3]{2}} \right)^3} = 3 - 2 = 1\]

b). Áp dụng hằng đẳng thức \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab(a + b)\]

Ta có: \[{B^3} = {\left( {\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}} \right)^3} = 2 + \sqrt 5  + 2 - \sqrt 5  + 3\sqrt[3]{{2 + \sqrt 5 }}\sqrt[3]{{2 - \sqrt 5 }}\left( {\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}} \right)\]

\[ = 4 + 3\sqrt[3]{{\left( {2 + \sqrt 5 } \right)\left( {2 - \sqrt 5 } \right)}}B = 4 + 3\sqrt[3]{{4 - 5}}B = 4 - 3B\]

\[\begin{array}{l}{B^3} + 3B - 4 = 0\\{B^3} - 1 + 3B - 3 = 0\end{array}\]

\[\left( {B - 1} \right)\left( {{B^2} + B + 4} \right) = 0\]

\[B = 1\, \left( {{B^2} + B + 4 = {B^2} + 2\frac{1}{2}B + \frac{1}{4} + \frac{{15}}{4} = {{\left( {B + \frac{1}{2}} \right)}^2} + \frac{{15}}{4} > 0} \right)\]

Vậy \[B = \sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }} = 1.\]