Đặt \(a=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(\Leftrightarrow a^3=\left(\sqrt[3]{2+\sqrt{5}}\right)^3+\left(\sqrt[3]{2-\sqrt{5}}\right)^3+3.\sqrt[3]{2+\sqrt{5}}.\sqrt[3]{2-\sqrt{5}}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(\Leftrightarrow a^3=2+\sqrt{5}+2-\sqrt{5}+3.\sqrt[3]{(2+\sqrt{5})\left(2-\sqrt{5}\right)}.a\)
\(\Leftrightarrow a^3=4+3.\sqrt[3]{4-5}.a\)
\(\Leftrightarrow a^3=4+3.\sqrt[3]{-1}.a\)
\(\Leftrightarrow a^3=4+3.\left(-1\right).a\)
\(\Leftrightarrow a^3=4-3.a\)
\(\Leftrightarrow a^3+3a-4=0\)
\(\Leftrightarrow\left(a-1\right)\left(a^2+a+4\right)=0\)
Mà \(a^2+a+4=\left(a+\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)
\(\Rightarrow a-1=0\)
\(\Leftrightarrow a=1\)
Vậy \(a=1\)
Áp dụng HĐT \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\), ta có:
\(A^3=2+\sqrt{5}+2-\sqrt{5}+3.A\sqrt[3]{4-5}\)
\(\Leftrightarrow A^3=4-3A\)
Giải phương trình: \(A^3+3A-4=0\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+1+4\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+5\right)=0\Leftrightarrow A=1\)
Vậy \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}=1}\)
