Đáp án:
\[P = 2017\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\left( {a + b} \right)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3} = \left( {{a^3} + {b^3}} \right) + 3ab.\left( {a + b} \right)\\
x = \sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}\\
\Leftrightarrow {x^3} = {\left( {\sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}} \right)^3}\\
\Leftrightarrow {x^3} = \left( {{{\sqrt[3]{{9 + 4\sqrt 5 }}}^3} + {{\sqrt[3]{{9 - 4\sqrt 5 }}}^3}} \right) + 3.\sqrt[3]{{9 + 4\sqrt 5 }}.\sqrt[3]{{9 - 4\sqrt 5 }}.\left( {\sqrt[3]{{9 + 4\sqrt 5 }} + \sqrt[3]{{9 - 4\sqrt 5 }}} \right)\\
\Leftrightarrow {x^3} = \left( {9 + 4\sqrt 5 + 9 - 4\sqrt 5 } \right) + 3.\sqrt[3]{{\left( {9 + 4\sqrt 5 } \right)\left( {9 - 4\sqrt 5 } \right)}}.x\\
\Leftrightarrow {x^3} = 18 + 3.\sqrt[3]{{{9^2} - {{\left( {4\sqrt 5 } \right)}^2}}}.x\\
\Leftrightarrow {x^3} = 18 + 3.\sqrt[3]{1}.x\\
\Leftrightarrow {x^3} - 3x = 18\\
y = \sqrt[3]{{3 + 2\sqrt 2 }} + \sqrt[3]{{3 - 2\sqrt 2 }}\\
\Leftrightarrow {y^3} = {\left( {\sqrt[3]{{3 + 2\sqrt 2 }} + \sqrt[3]{{3 - 2\sqrt 2 }}} \right)^3}\\
\Leftrightarrow {y^3} = \left( {{{\sqrt[3]{{3 + 2\sqrt 2 }}}^3} + {{\sqrt[3]{{3 - 2\sqrt 2 }}}^3}} \right) + 3.\sqrt[3]{{3 + 2\sqrt 2 }}.\sqrt[3]{{3 - 2\sqrt 2 }}.\left( {\sqrt[3]{{3 + 2\sqrt 2 }} + \sqrt[3]{{3 - 2\sqrt 2 }}} \right)\\
\Leftrightarrow {y^3} = \left( {3 + 2\sqrt 2 + 3 - 2\sqrt 2 } \right) + 3.\sqrt[3]{{\left( {3 + 2\sqrt 2 } \right).\left( {3 - 2\sqrt 2 } \right)}}.y\\
\Leftrightarrow {y^3} = 6 + 3.\sqrt[3]{{{3^2} - {{\left( {2\sqrt 2 } \right)}^2}}}.y\\
\Leftrightarrow {y^3} = 6 + 3.1.y\\
\Leftrightarrow {y^3} - 3y = 6\\
P = {x^3} + {y^3} - 3\left( {x + y} \right) + 1993\\
= \left( {{x^3} - 3x} \right) + \left( {{y^3} - 3y} \right) + 1993\\
= 18 + 6 + 1993\\
= 2017
\end{array}\)
Vậy \(P = 2017\)
