Đặt B=.......
Lập phương 2 vế:
[tex]B^3=12+3\sqrt[3]{6+\frac{\sqrt{247}}{\sqrt{27}}}.\sqrt[3]{6-\frac{\sqrt{247}}{\sqrt{27}}}.(\sqrt[3]{6+\frac{\sqrt{247}}{\sqrt{27}}}+\sqrt[3]{6-\frac{\sqrt{247}}{\sqrt{27}}})=12+3.\frac{5}{3}.B<=>B^3-5B-12=0<=>B=3[/tex]