Question

∛(10+√68) · ∛(10-√68) ?

Answer 1

            
$Folosim \,\,\,formulele: \\ \sqrt[n]{a} * \sqrt[n]{b}= \sqrt[n]{a*b} \\ si \\ (a+b)(a-b)= a^{2}- b^{2} \\ \\ Rezolvare: \\ \\ \sqrt[3]{(10+ \sqrt{68})} * \sqrt[3]{(10- \sqrt{68})}= \sqrt[3]{(10+ \sqrt{68})(10- \sqrt{68})} = \\ = \sqrt[3]{(10^{2} - (\sqrt{68})^{2})}= \sqrt[3]{(100 - 68)} =\sqrt[3]{32}=2\sqrt[3]{4}$