Question

Cho \(a=\)\(\sqrt[3]{3+\sqrt{17}}+\sqrt[3]{3-\sqrt{17}}\). Tính A = (\(a^3+6a-5\))\(^{2019}\)

Answer 1

Ta có: \(a^3=\left(\sqrt[3]{3+\sqrt{17}}+\sqrt[3]{3-\sqrt{17}}\right)^3\)

\(=3+\sqrt{17}+3-\sqrt{17}+3\sqrt[3]{\left(3+\sqrt{17}\right)\left(3-\sqrt{17}\right)}\left(\sqrt[3]{3+\sqrt{17}}+\sqrt[3]{3-\sqrt{17}}\right)\)

(\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\) )

\(=6+3\sqrt[3]{-8}.a=6-6a\)

\(\Rightarrow a^3+6a-6=0\Rightarrow a^3+6a-5=1\)

\(\Rightarrow A=1^{2019}=1\)