# How do you simplify root4(x^12/y^4)?

Feb 25, 2017

$\sqrt[4]{{x}^{12} / {y}^{4}} = {x}^{3} / y$

#### Explanation:

$\sqrt[4]{{x}^{12} / {y}^{4}} = {\left({x}^{12} / {y}^{4}\right)}^{\frac{1}{4}} = {\left({x}^{12} \cdot {y}^{-} 4\right)}^{\frac{1}{4}}$

Remember the rule of indices: ${\left({a}^{m}\right)}^{n} = {a}^{m \times n}$

Applying this rule to the expression:

${\left({x}^{12} \cdot {y}^{-} 4\right)}^{\frac{1}{4}} = {x}^{\frac{12}{4}} \cdot {y}^{- \frac{4}{4}}$

$= {x}^{3} \cdot {y}^{-} 1$

$= {x}^{3} / y$

Feb 25, 2017

Detailed explanation:

#### Explanation:

color(blue)(root(4)((x^12)/(y^4))

Let's solve it using simple steps

First we should know that color(brown)(root(x)(y/z)=(root(x)(y))/(root(x)(z))

So,

$\rightarrow \sqrt[4]{\frac{{x}^{12}}{{y}^{4}}} = \frac{\sqrt[4]{{x}^{12}}}{\sqrt[4]{{y}^{4}}}$

Now solve $\sqrt[4]{{x}^{12}}$ and $\sqrt[4]{{y}^{4}}$ each independantly

Let's solve $\sqrt[4]{{x}^{12}}$ (expand it)

$\rightarrow \sqrt[4]{{x}^{12}} = \sqrt[4]{x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}$

Take out the roots

$\rightarrow \sqrt[4]{\underbrace{x \cdot x \cdot x \cdot x} \cdot \underbrace{x \cdot x \cdot x \cdot x} \cdot \underbrace{x \cdot x \cdot x \cdot x}}$

$\rightarrow x \cdot x \cdot x$

color(green)(rArrx^3

Now solve $\sqrt[4]{{y}^{4}}$ (expand it)

$\rightarrow \sqrt[4]{{y}^{4}} = \sqrt[4]{y \cdot y \cdot y \cdot y}$

Take out the roots

$\rightarrow \sqrt[4]{\underbrace{y \cdot y \cdot y \cdot y}}$

color(green)(rArry

So, put the solutions together to get our answer

color(blue)(x^3/y

Hope this helps! :)