Verified answer

[tex]\displaystyle\bf\\a=\Big(12-\sqrt{80} \Big)^{\frac{1}{3} }\sqrt[3]{12+80^{0,5} } =\sqrt[3]{12-\sqrt{80} } \cdot\sqrt[3]{12+\sqrt{80} } =\\\\\\=\sqrt[3]{(12-\sqrt{80} )\cdot(12+\sqrt{80} )} =\sqrt[3]{12^{2} -(\sqrt{80})^{2} } =\\\\\\=\sqrt[3]{144-80} =\sqrt[3]{64} =\sqrt[3]{4^{3} } =4\\\\\\Otvet \ : \ a=4[/tex]