Đặt A= \(\sqrt[3]{3+2\sqrt{2}}\) + \(\sqrt[3]{3-2\sqrt{2}}\)
=> \(A^3\) = \(3+2\sqrt{2}\) + \(3-2\sqrt{2}\) + 3.\(\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\) \(\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)\)
= 6 + 3. \(\sqrt[3]{3^2-\left(2\sqrt{2}\right)^2}\). A
=6 + 3A => Tính đc A ( số ko đẹp lắm !!)