Answer
$a)$ $\sqrt{y}$
$b)$ $\dfrac{4u}{v^{2}}$
Work Step by Step
$a)$ $\sqrt[3]{y\sqrt{y}}$
Rewrite the roots as powers with rational exponent:
$\sqrt[3]{y\sqrt{y}}=(y\cdot y^{1/2})^{1/3}=...$
Evaluate the product inside the parentheses:
$...=(y^{3/2})^{1/3}=...$
Finally, evaluate the power:
$...=y^{1/2}=\sqrt{y}$
$b)$ $\sqrt{\dfrac{16u^{3}v}{uv^{5}}}$
Evaluate the division inside the square root:
$\sqrt{\dfrac{16u^{3}v}{uv^{5}}}=\sqrt{16u^{3-1}v^{1-5}}=\sqrt{16u^{2}v^{-4}}=...$
Evaluate the square root and simplify if possible:
$...=4uv^{-2}=\dfrac{4u}{v^{2}}$
