Radicales necesito resolver este ejercicio de radicales

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$$\begin{align}&Primero\ simplificaremos\ el\ numerador:\\ \\&\frac{5\sqrt[3] {297}-2\sqrt[3] {189}}{3\sqrt[3] {2}} \\\\&\frac{5\sqrt[3] {3*3*3*11}-2\sqrt[3] {3*3*3*7}}{3\sqrt[3] {2}} \\\\&\frac{5\sqrt[3] {3^3*11}-2\sqrt[3] {3^3*7}}{3\sqrt[3] {2}} \\\\&\frac{5\sqrt[3] {3^3}\sqrt[3] {11}-2\sqrt[3] {3^3}\sqrt[3] {7}}{3\sqrt[3] {2}} \\\\&\frac{5*3\sqrt[3] {11}-2*3\sqrt[3] {7}}{3\sqrt[3] {2}} \\\\&\frac{15\sqrt[3] {11}-6\sqrt[3] {7}}{3\sqrt[3] {2}} \\\\&Ahora\ Racioalizaremos\ el\ denominador:\\\\&\frac{15\sqrt[3] {11}-6\sqrt[3] {7}}{3\sqrt[3] {2}}*\frac{\sqrt[3] {2^2}}{\sqrt[3] {2^2}}\\\\&\frac{(15\sqrt[3] {11}-6\sqrt[3] {7})(\sqrt[3] {2^2})}{3\sqrt[3] {2}*\sqrt[3] {2^2}}\\\\&\frac{15\sqrt[3] {11}*\sqrt[3] {2^2}-6\sqrt[3] {7}*\sqrt[3] {2^2}}{3\sqrt[3] {2^3}}\\\\&\frac{15\sqrt[3] {11}*\sqrt[3] {4}-6\sqrt[3] {7}*\sqrt[3] {4}}{3*2}\\\\&\frac{15\sqrt[3] {11*4}-6\sqrt[3] {7*4}}{3*2}\\\\&\frac{3(5\sqrt[3] {44}-2\sqrt[3] {28})}{3*2}\\\\\\&\frac{5\sqrt[3] {44}-2\sqrt[3] {28}}{2}\\\\&\end{align}$$