[tex]\text{ }\sqrt[3]{\frac{1}{3}}\text{ (option A)}[/tex]Explanation:[tex]\begin{gathered} 2)\text{ }\frac{\sqrt[3]{8}}{\sqrt[3]{24}} \\ \\ We\text{ will solve numerator and denominator seperately. Then we will combine the results} \\ 8=2^3 \\ \sqrt[3]{8}=\text{ }\sqrt[3]{2^3}=\text{ 2} \end{gathered}[/tex][tex]\begin{gathered} 24\text{ = 3 }\times\text{ 8} \\ \sqrt[3]{24}\text{ = }\sqrt[3]{3\times8}\text{ = }\sqrt[3]{3}\text{ }\times\text{ }\sqrt[3]{8} \\ \sqrt[3]{24}\text{ = }\sqrt[3]{3}\text{ }\times\text{ }\sqrt[3]{2^3} \\ \sqrt[3]{24}\text{ = }\sqrt[3]{3}\text{ }\times\text{ 2} \end{gathered}[/tex][tex]\begin{gathered} \frac{\sqrt[3]{8}}{\sqrt[3]{24}}\text{ = }\frac{2}{\sqrt[3]{3}\text{ }\times\text{ 2}} \\ 2\text{ cancles out:} \\ \\ \frac{\sqrt[3]{8}}{\sqrt[3]{24}}\text{ = }\frac{1}{\sqrt[3]{3}\text{ }} \end{gathered}[/tex][tex]\begin{gathered} \sin ce\text{ we don't have the result above in the question, we can simplify further} \\ By\text{ re-w riting in another form:} \\ \frac{\sqrt[3]{8}}{\sqrt[3]{24}}\text{ = }\frac{1}{\sqrt[3]{3}\text{ }} \\ \sqrt[3]{1}\text{ = 1} \\ \\ \frac{\sqrt[3]{8}}{\sqrt[3]{24}}\text{ = }\frac{\sqrt[3]{1}}{\sqrt[3]{3}\text{ }}\text{ cube root is common to the numerator and denominator},\text{ combine} \\ \\ \text{= }\sqrt[3]{\frac{1}{3}}\text{ (option A)} \end{gathered}[/tex]
