[tex]x=1+\sqrt[3]{2}+\sqrt[3]{4}\Rightarrow (\sqrt[3]{2}-1)x=(\sqrt[3]{2}-1)(1+\sqrt[3]{2}+\sqrt[3]{4})=1\Rightarrow \frac{1}{x}=\sqrt[3]{2}-1\Rightarrow \frac{x+1}{x}=\sqrt[3]{2}\Rightarrow \frac{(x+1)^3}{x^3}=2\Rightarrow x^3+3x^2+3x+1=2x^3\Leftrightarrow x^3-3x^2-3x=1\Rightarrow A=\sqrt{2020}[/tex]