Đáp án:
$\begin{array}{l}
A = \sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}} + \sqrt[3]{{1 - \dfrac{{\sqrt {84} }}{9}}}\\
\Leftrightarrow {A^3} = 1 + \dfrac{{\sqrt {84} }}{9} + 3{\left( {\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}} \right)^2}.\sqrt[3]{{1 - \dfrac{{\sqrt {84} }}{9}}}\\
+ 3.\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}.{\left( {\sqrt[3]{{1 - \dfrac{{\sqrt {84} }}{9}}}} \right)^2} + 1 - \dfrac{{\sqrt {84} }}{9}\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}.\sqrt[3]{{1 - \dfrac{{\sqrt {84} }}{9}}}.\left( {\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}} - \sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}} \right)\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{1 - \dfrac{{84}}{{81}}}}.A\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{\dfrac{{ - 3}}{{81}}}}.A\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{ - \dfrac{1}{{27}}}}.A\\
\Leftrightarrow {A^3} = 2 + 3.\dfrac{{ - 1}}{3}.A\\
\Leftrightarrow {A^3} + A - 2 = 0\\
\Leftrightarrow {A^3} - {A^2} + {A^2} - A + 2A - 2 = 0\\
\Leftrightarrow \left( {A - 1} \right)\left( {{A^2} + A + 2} \right) = 0\\
\Leftrightarrow A = 1
\end{array}$
Vậy A có giá trị là 1 số nguyên
