Đặt [tex]\sqrt[3]{4}=x,\sqrt[3]{y}\Rightarrow \sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9}=x^2+xy+y^2\Rightarrow \frac{1}{\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9}}=\frac{1}{x^2+xy+y^2}=\frac{x-y}{(x-y)(x^2+xy+y^2)}=\frac{x-y}{x^3-y^3}=\frac{\sqrt[3]{4}-\sqrt[3]{3}}{4-3}=\sqrt[3]{4}-\sqrt[3]{3}[/tex]