# How do you simplify sqrt(x^3y^9)?

First of all, use the fact that $\setminus \sqrt{a \setminus \cdot b} = \setminus \sqrt{a} \setminus \sqrt{b}$. So, we have that $\setminus \sqrt{{x}^{3} {y}^{9}} = \setminus \sqrt{{x}^{3}} \setminus \sqrt{{y}^{9}}$
Let's deal with a root at a time: we can write ${x}^{3}$ as ${x}^{2} \setminus \cdot x$. So, $\setminus \sqrt{{x}^{3}} = \setminus \sqrt{{x}^{2} \setminus \cdot x} = \setminus \sqrt{{x}^{2}} \setminus \sqrt{x} = x \setminus \sqrt{x}$.
For the same reasons, we write ${y}^{9}$ as ${y}^{2} \setminus \cdot {y}^{2} \setminus \cdot {y}^{2} \setminus \cdot {y}^{2} \setminus \cdot y$. So, we have that $\setminus \sqrt{{y}^{9}} = \setminus \sqrt{{y}^{2} \setminus \cdot {y}^{2} \setminus \cdot {y}^{2} \setminus \cdot {y}^{2} \setminus \cdot y} = \setminus \sqrt{{y}^{2}} \setminus \sqrt{{y}^{2}} \setminus \sqrt{{y}^{2}} \setminus \sqrt{{y}^{2}} \setminus \sqrt{y}$, which equals ${y}^{4} \setminus \sqrt{y}$.
$\setminus \sqrt{{x}^{3} {y}^{9}} = x {y}^{4} \setminus \sqrt{x y}$.