Trục căn ở mẫu số biểu diễn \(\frac{1}{{\sqrt[3]{{16}} + \sqrt[3]{{12}} + \sqrt[3]{9}}}\).

a) Ta có \(A = \left[ {\left( {\frac{1}{a} - \sqrt[6]{{\frac{1}{a}}} + \sqrt[3]{{{a^2}}}} \right) + \left( {\frac{a}{{{a^2}}}\sqrt[6]{{{a^5}}} - \frac{3}{a}\sqrt[3]{{{a^2}}}} \right)} \right]a\sqrt[3]{a}\)

\( = \left[ {\left( {\frac{1}{a} - \sqrt[6]{{\frac{1}{a}}} + \sqrt[3]{{{a^2}}}} \right) + \left( {\sqrt[6]{{\frac{1}{a}}} - 3\sqrt[3]{{\frac{1}{a}}}} \right)} \right]a\sqrt[3]{a}\)

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

\( = {a^2} - 3a + \sqrt[3]{a}\).

b)  Ta có \(B = \sqrt[3]{{{a^2}}}\sqrt {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}}  + \sqrt[3]{{{b^2}}}\sqrt {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}} \)

\( = \left( {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}} \right)\sqrt {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}} \)

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

a) \(\sqrt[3]{{64}} = 4\).

b) \(\sqrt[3]{{ - 512}} =  - 8\)

c) \(\sqrt[3]{{0,064}} = \sqrt[3]{{\frac{{64}}{{1000}}}} = \frac{{\sqrt[3]{{64}}}}{{\sqrt[3]{{1000}}}} = \frac{4}{{10}}\).

d) \(\sqrt[3]{{ - 0,216}} = \sqrt[3]{{{{\left( { - 6} \right)}^3}}} =  - 6\).

e) \(\frac{{\sqrt[3]{{500}}}}{{\sqrt[3]{4}}} + \sqrt[3]{{12}} \cdot \sqrt[3]{{18}} = \sqrt[3]{{\frac{{500}}{4}}} + \sqrt[3]{{12 \cdot 18}} = \sqrt[3]{{125}} + \sqrt[3]{{216}} = 5 + 6 = 11\)

f) \(\frac{{\sqrt[3]{{12}} \cdot \sqrt[3]{6}}}{{\sqrt[3]{{576}}}} - \frac{{\sqrt[3]{{32}}}}{{\sqrt[3]{4}}} = \sqrt[3]{{\frac{{12 \cdot 6}}{{576}}}} - \sqrt[3]{{\frac{{32}}{4}}} = \sqrt[3]{{\frac{1}{8}}} - \sqrt[3]{8} = \frac{1}{2} - 2 =  - \frac{3}{2}\).