Odpowiedź:
[tex]a)\ \ (\frac{1}{2}\sqrt[3]{12})^3=(\frac{1}{2})^3\cdot(\sqrt[3]{12})^3=\frac{1}{\not8_{2} }\cdot\not12^3=\frac{1}{2}\cdot3 =\frac{3}{2}=1\frac{1}{2}\\\\\\b)\ \ \left(-2\sqrt[3]{\frac{1}{4}}\right)^3=(-2)^3\cdot\left(\sqrt[3]{\frac{1}{4}}\right)^3=-\not8^2\cdot\frac{1}{\not4_{1}}=-2\\\\\\c)\ \ (-10\sqrt[3]{-0,1})^3=(-10\cdot(-\sqrt[3]{0,1}))^3=(10\sqrt[3]{0,1})^3=10^3\cdot(\sqrt[3]{0,1})^3=\\\\=1000\cdot0,1=100[/tex]
[tex]d)\ \ 3\sqrt[3]{7}\cdot2(\sqrt[3]{7})^2=3\sqrt[3]{7}\cdot2\sqrt[3]{49}=3\cdot2\sqrt[3]{7\cdot49}=6\sqrt[3]{7\cdot7^2}=6\sqrt[3]{7^3}=6\cdot7=42\\\\\\e)\ \ 5(\sqrt[3]{-6})^2\cdot4\sqrt[3]{-6}=5\sqrt[3]{36}\cdot4(-\sqrt[3]{6})=-5\sqrt[3]{36}\cdot4\sqrt[3]{6}=-5\cdot4\sqrt[3]{36\cdot6}=\\\\=-20\sqrt[3]{6^2\cdot6}=-20\sqrt[3]{6^3}=-20\cdot6=-120[/tex]
[tex]f)\ \ (-7\sqrt[3]{-2})^2\cdot\sqrt[3]{-2}=(7\sqrt[3]{2})^2\cdot(-\sqrt[3]{2})=7^2\cdot(\sqrt[3]{2})^2\cdot(-\sqrt[3]{2})=-49\cdot\sqrt[3]{4}\cdot\sqrt[3]{2}=\\\\=-49\sqrt[3]{4\cdot2}=-49\sqrt[3]{8}=-49\cdot2=-98[/tex]
