# How do you write the expression sqrt(48x^6y^3) in simplified radical form.

First of all, let's deal with the numeric part: by factoring 48 with prime numbers, $\setminus \sqrt{48} = \setminus \sqrt{{2}^{4} \setminus \cdot 3}$. Now use the fact that the square root of a product is the product of the square roots, so we have $\setminus \sqrt{48} = \setminus \sqrt{16} \setminus \cdot \setminus \sqrt{3} = 4 \setminus \sqrt{3}$
As for the variables, we can again say that $\setminus \sqrt{{x}^{6} {y}^{3}} = \setminus \sqrt{{x}^{6}} \setminus \sqrt{{y}^{3}}$. You can write ${x}^{6}$ as ${x}^{2} \setminus \cdot {x}^{2} \setminus \cdot {x}^{2}$, so $\setminus \sqrt{{x}^{6}} = \setminus \sqrt{{x}^{2} \setminus \cdot {x}^{2} \setminus \cdot {x}^{2}} = \setminus \sqrt{{x}^{2}} \setminus \cdot \setminus \sqrt{{x}^{2}} \setminus \cdot \setminus \sqrt{{x}^{2}} = x \setminus \cdot x \setminus \cdot x = {x}^{3}$
The same reason bring us to write ${y}^{3}$ as ${y}^{2} \setminus \cdot y$, so $\setminus \sqrt{{y}^{3}} = \setminus \sqrt{{y}^{2} \setminus \cdot y} = \setminus \sqrt{{y}^{2}} \setminus \cdot \setminus \sqrt{y} = y \setminus \sqrt{y}$.
$\setminus \sqrt{48 {x}^{6} {y}^{3}} = 4 {x}^{3} y \setminus \sqrt{3 y}$