## Precalculus (6th Edition) Blitzer

$y\sqrt{y}$
RECALL: (i) For any non-negative real numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. (ii) For any non-negative real number $a$, $\sqrt{a^2}=a$. Factor the radicand so that one of the factors is a perfect square to obtain $\\\sqrt{y^2(y)}.$ Use rule (i) above to obtain $\sqrt{y^2} \cdot \sqrt{y}.$ Use rule (ii) above to obtain $y\sqrt{y}.$