Ta có \(\sqrt[3]{26+15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}\)
\(=\sqrt[3]{2^3+3.2^2\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}\)
Đặt \(x=\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)
Ta có \(x^3=\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)^3\)
\(=9+\sqrt{80}+9-\sqrt{80}+3.\left(\sqrt[3]{9+\sqrt{80}}\right)^2\left(\sqrt[3]{9-\sqrt{80}}\right)+3.\left(\sqrt[3]{9-\sqrt{80}}\right)^2\left(\sqrt[3]{9+\sqrt{80}}\right)\)
\(=18+3\sqrt[3]{9+\sqrt{80}}.\sqrt[3]{9-\sqrt{80}}\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)\)
\(=18+3\sqrt[3]{9^2-80}.x\)
\(=18+3x\)
Vậy \(x^3=18+3x\)
\(\Leftrightarrow x^3-3x-18=0\)
Vậy x = 3
Do đó \(M=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+3=2^2-3+3=4\)
Vậy M = 4.
