Đáp án:
\[f\left( x \right) = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
x = \sqrt[3]{{16 - 8\sqrt 5 }} + \sqrt[3]{{16 + 8\sqrt 5 }}\\
\Leftrightarrow {x^3} = {\left( {\sqrt[3]{{16 - 8\sqrt 5 }} + \sqrt[3]{{16 + 8\sqrt 5 }}} \right)^3}\\
\Leftrightarrow {x^3} = {\sqrt[3]{{16 - 8\sqrt 5 }}^3} + 3.{\sqrt[3]{{16 - 8\sqrt 5 }}^2}.\sqrt[3]{{16 + 8\sqrt 5 }} + 3.\sqrt[3]{{16 - 8\sqrt 5 }}.{\sqrt[3]{{16 + 8\sqrt 5 }}^2} + {\sqrt[3]{{16 + 8\sqrt 5 }}^3}\\
\Leftrightarrow {x^3} = \left( {16 - 8\sqrt 5 } \right) + 3.\sqrt[3]{{16 - 8\sqrt 5 }}.\sqrt[3]{{16 + 8\sqrt 5 }}.\left( {\sqrt[3]{{16 - 8\sqrt 5 }} + \sqrt[3]{{16 + 8\sqrt 5 }}} \right) + \left( {16 + 8\sqrt 5 } \right)\\
\Leftrightarrow {x^3} = 16 - 8\sqrt 5 + 3.\sqrt[3]{{\left( {16 - 8\sqrt 5 } \right)\left( {16 + 8\sqrt 5 } \right)}}.x + 16 + 8\sqrt 5 \\
\Leftrightarrow {x^3} = 32 + 3.\sqrt[3]{{{{16}^2} - {{\left( {8\sqrt 5 } \right)}^2}}}.x\\
\Leftrightarrow {x^3} = 32 + 3.\sqrt[3]{{ - 64}}.x\\
\Leftrightarrow {x^3} = 32 + 3.\left( { - 4} \right).x\\
\Leftrightarrow {x^3} + 12x - 32 = 0\\
\Rightarrow {x^3} + 12x - 31 = 1\\
\Rightarrow f\left( x \right) = {\left( {{x^3} + 12x - 31} \right)^{2021}} = {1^{2021}} = 1
\end{array}\)
Vậy \(f\left( x \right) = 1\)
