Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)
Ta đặt: \(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}.x\)
\(=18+3\sqrt[3]{81-80}.x\)
\(=18+3x\)
\(\Rightarrow x^3-18-3x=0\)
\(\Rightarrow x^3-3x^2+3x^2-9x+6x-18=0\)
\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+6\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+6\right)=0\)
Vì \(x^2+3x+6=x^2+2.x.\frac{3}{2}+\frac{9}{4}+\frac{15}{4}=\left(x+\frac{3}{2}\right)^2+\frac{15}{4}>0\)
Suy ra: x - 3 = 0
=> x = 3
Vâỵ \(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}=3\)
