Simplifying a radical equation?

Question

I'm lost on how to simplify this.

Simplify: $$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}}$$

OK So I think I got it.

$$\frac{\sqrt[\large 4]{144x^9y^8}}{\sqrt[\large 4]{9x^5y^{-3}}} = \left(\frac{{144x^9y^8}}{{9x^5y^{-3}}}\right )^{1/4}$$

You can break that down like this -I think-:

$$\sqrt[\large 4]{16x^4y^{11}}$$

And then finally:

$$2xy^2\sqrt[\large 4]{y^3}$$

EDIT - You can go even further (Thanks amWhy):

$$2xy^{\large \frac{11}{4}}$$

Answer 1

First, recall that

$$ \frac{\sqrt[\large4]{144x^9y^8}}{\sqrt[\large4]{9x^5y^{-3}}} = \sqrt[\large 4]{\frac{144x^9y^8}{9x^5y^{-3}}} = \left(\frac{144x^9y^8}{9x^5y^{-3}}\right)^{1/4}$$

Now, use the "laws" of exponents to simplify the rational expression, (the fraction inside parentheses), and then take distribute the exponent $1/4$ over the factors remaining after simplifying.

ADDED: You are almost there: you can also write the exponent of $y$ as a fraction. Recall that $\;\;(y^a)^b = y^{ab}.\;$ In your case, $\,a = 3$, and $\;b=1/4$:

$$2xy^2\sqrt[4]{y^3} = 2xy^2\left((y^3)^{1/4}\right) = 2x y^2 \left(y^{3/4}\right) = 2x\left(\color{blue}{\bf y^{\left(2 + \frac 34\right)}}\right)\quad ?$$

$\color{blue}{\bf \left(y^cy^d = y^{(c + d)}\right)}$

Answer 2

Hint: Start by using the fact that $$\frac{\sqrt[4]{a}}{\sqrt[4]{b}}=\sqrt[4]{\frac ab},$$ and see what you can do on the inside of the radical. Then use the fact that $\sqrt[4]c=c^{1/4}$ with exponent rules. Don't forget that $16=2^4.$

Edit: Bingo! Nicely done.