\(A=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)
\(\Rightarrow A^3=14+3\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\)
<=>A3=14-3A
<=>A3+3A-14=0
<=>A3-4A+7A-14=0
<=>A.(A-2)(A+2)+7.(A-2)=0
<=>(A-2)(A2+9A-14)=0
<=>A=2(nhận)
Vậy A=2