Đặt \(A=\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}\)
Ta có: \(A^3=\left(\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}\right)^3\)
\(=99-70\sqrt{2}+99+70\sqrt{2}+3\cdot\sqrt[3]{\left(99-70\sqrt{2}\right)\left(99+70\sqrt{2}\right)}\cdot A\)
\(=198+3A\)
\(\Leftrightarrow A^3-198-3A=0\)
\(\Leftrightarrow A^3-3A-198=0\)
\(\Leftrightarrow A^3-6A^2+6A^2-36A+33A-198=0\)
\(\Leftrightarrow A^2\left(A-6\right)+6A\left(A-6\right)+33\left(A-6\right)=0\)
\(\Leftrightarrow\left(A-6\right)\left(A^2+6A+33\right)=0\)
mà \(A^2+6A+33>0\forall A\)
nên A-6=0
hay A=6
Vậy: \(\sqrt[3]{99-70\sqrt{2}}+\sqrt[3]{99+70\sqrt{2}}=6\)
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