Exercise ID91

Algebra → Roots → Simplifying $$\left(\sqrt[3]{1+2\sqrt{6}}-\sqrt[6]{25+4\sqrt{6}}\right)·\sqrt[3]{2\sqrt{6}-1}$$ $$=\sqrt[3]{1+2\sqrt{6}}·\sqrt[3]{2\sqrt{6}-1}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[3]{2\sqrt{6}-1}$$ $$=\sqrt[3]{1+2\sqrt{6}}·\sqrt[3]{2\sqrt{6}-1}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[3]{2\sqrt{6}-1}$$ $$=\sqrt[3]{{\left(2\sqrt{6}\right)}^2-1^{{}^2}}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[3]{2\sqrt{6}-1}$$ $$=\sqrt[3]{23}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[3]{2\sqrt{6}-1}$$ $$=\sqrt[3]{23}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[6]{{\left(2\sqrt{6}-1\right)}^2}$$ $$=\sqrt[3]{24-1}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[6]{24-4\sqrt{6}+1}$$ $$=\sqrt[3]{23}-\sqrt[6]{25+4\sqrt{6}}·\sqrt[6]{25-4\sqrt{6}}$$ $$=\sqrt[3]{23}-\sqrt[6]{25+4\sqrt{6}·}\sqrt[6]{25-4\sqrt{6}}$$ $$=\sqrt[3]{23}-\sqrt[6]{{25}^2-{\left(4\sqrt{6}\right)}^2}$$ $$=\sqrt[3]{23}-\sqrt[6]{625-96}$$ $$=\sqrt[3]{23}-\sqrt[6]{529}$$ $$=\sqrt[3]{23}-\sqrt[6]{{23}^2}$$ $$=\sqrt[3]{23}-\sqrt[3]{23}$$ $$=0$$