## College Algebra 7th Edition

$\textbf{(a)} \hspace{0.7cm}3\sqrt[3]{2}$ $\textbf{(b)}\hspace{0.7cm}\sqrt{10}(\sqrt{5}+2)$ $\textbf{(c)}\hspace{0.7cm}\dfrac{1-\sqrt{x}}{1-x}$
$\textbf{(a)} \hspace{0.7cm}\dfrac{6}{\sqrt[3]{4}}=\dfrac{6}{\sqrt[3]{4}}.\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}}=\dfrac{6\sqrt[3]{2}}{\sqrt[3]{8}}=\dfrac{6\sqrt[3]{2}}{2}=3\sqrt[3]{2}$ $\textbf{(b)} \hspace{0.7cm}\dfrac{\sqrt{10}}{\sqrt{5}-2}=\dfrac{\sqrt{10}}{\sqrt{5}-2}.\dfrac{\sqrt{5}+2}{\sqrt{5}+2}=\dfrac{\sqrt{10}(\sqrt{5}+2)}{5-4}=\sqrt{10}(\sqrt{5}+2)$ $\textbf{(c)} \hspace{0.7cm}\dfrac{1}{1+\sqrt{x}}=\dfrac{1}{1+\sqrt{x}}.\dfrac{1-\sqrt{x}}{1-\sqrt{x}}=\dfrac{1-\sqrt{x}}{1-x}$