Giải phương trình:

a) \(\sqrt[3]{{{x^3} + 9{x^2}}} = x + 3\)                                                              b) \(\sqrt[3]{{5 + x}} - x = 5.\)

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

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

b) \({B^3} = 182 + \sqrt {33125}  + 182 - \sqrt {33125}  + 3\sqrt[3]{{{{182}^2} - 33125}}B = 364 - 3B.\)

Khi đó  \({B^3} + 3B - 364 = 0\) nên \(\left( {B - 7} \right)\left( {{B^2} + 7B + 52} \right) = 0\), suy ra \(B = 7\).

(do \({B^2} + 7B + 52 = {\left( {B + \frac{7}{2}} \right)^2} + \frac{{159}}{4} > 0\)).

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

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

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