Đặt \(A=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)
Áp dụng hằng đẳng thức \(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\)ta có:
\(A^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(=\left(7+5\sqrt{2}\right)+\left(7-5\sqrt{2}\right)+\)\(3\sqrt[3]{7+5\sqrt{2}}\cdot\sqrt[3]{7-5\sqrt{2}}\cdot\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)
\(=14+3\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\cdot A\)
\(=14+3\sqrt[3]{49-50}\cdot A\)
\(=14+3\sqrt[3]{-1}\cdot A\)
\(=14-3A.\)
\(\Rightarrow A^3+3A-14=0\)
\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+7\right)=0.\)
Ta thấy rằng \(A^2+2A+7=\left(A+1\right)^2+6>0\)nên từ phương trình trên suy ra \(A-2=0\Rightarrow A=2.\)
Vậy A = 2.
