Theo giả thiết, ta có:
- \(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)^3\)
\(=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{9+4\sqrt{5}}.\sqrt[3]{9-4\sqrt{5}}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3.\sqrt[3]{1}.x\)
\(=18+3x\)
- \(y=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
\(\Rightarrow y^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{1}.y\)
\(=6+3y\)
Do đó: \(P=x^3+y^3-3\left(x+y\right)+1993\)
\(=18+3x+6+3y-3x-3y+1993\)
\(=2017\)
