Tính các căn bậc ba sau

a) \(\sqrt[3]{{64}}\)           b) \(\sqrt[3]{{ - 512}}\).                                            c) \(\sqrt[3]{{0,064}}\)

d) \(\sqrt[3]{{ - 0,216}}\)  e) \(\frac{{\sqrt[3]{{500}}}}{{\sqrt[3]{4}}} + \sqrt[3]{{12}} \cdot \sqrt[3]{{18}}\)                 f) \(\frac{{\sqrt[3]{{12}} \cdot \sqrt[3]{6}}}{{\sqrt[3]{{576}}}} - \frac{{\sqrt[3]{{32}}}}{{\sqrt[3]{4}}}\)

a) \[\sqrt[3]{{ - 64}} - \sqrt[3]{{125}} + \sqrt[3]{{216}} =  - 4 - 5 + 6 =  - 3\]

b) Ta có \[{\left( {\sqrt[3]{4} + 1} \right)^3} - {\left( {\sqrt[3]{4} - 1} \right)^3}\]

\[ = \left( {\sqrt[3]{4} + 1 - \sqrt[3]{4} + 1} \right)\left[ {{{\left( {\sqrt[3]{4} + 1} \right)}^2} + \left( {\sqrt[3]{4} + 1} \right)\left( {\sqrt[3]{4} - 1} \right) + {{\left( {\sqrt[3]{4} - 1} \right)}^2}} \right]\]

\[ = 2.\left( {\sqrt[3]{{16}} + 2\sqrt[3]{4} + 1 + \sqrt[3]{{16}} - 1 + \sqrt[3]{{16}} - 2\sqrt[3]{4} + 1} \right) = 2\left( {3\sqrt[3]{{16}} + 1} \right) = 2\left( {6\sqrt[3]{2} + 1} \right)\]

c) Ta có \[\left( {12\sqrt[3]{2} + \sqrt[3]{{16}} - 2\sqrt[3]{2}} \right)\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right)\]

\[ = \left( {12\sqrt[3]{2} + 2\sqrt[3]{2} - 2\sqrt[3]{2}} \right)\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right) = 12\sqrt[3]{2}\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right) = 120 - 36 = 84\]

Ta có \[x = \frac{2}{{2\sqrt[3]{2} + 2 + \sqrt[3]{4}}} = \frac{2}{{\sqrt[3]{{16}} + 2 + \sqrt[3]{4}}}\]

\[ = \frac{2}{{{{\left( {\sqrt[3]{4}} \right)}^2} + \sqrt[3]{2}.\sqrt[3]{4} + {{\left( {\sqrt[3]{2}} \right)}^2}}}\]

\[ = \frac{{2\left( {\sqrt[3]{4} - \sqrt[3]{2}} \right)}}{{4 - 2}} = \sqrt[3]{4} - \sqrt[3]{2}\]

Tương tự \(y = \frac{6}{{2\sqrt[3]{2} - 2 + \sqrt[3]{4}}} = \sqrt[3]{4} + \sqrt[3]{2}\)

Do đó: \(x{y^3} - {x^3}y = xy\left( {{y^2} - {x^2}} \right) = xy\left( {y - x} \right)\left( {y + x} \right) = 8\left( {2\sqrt[3]{2} - \sqrt[3]{4}} \right)\)