# How do you use the laws of exponents to simplify the expression  ((3^2)/(3^-3))^(3/5)?

Mar 28, 2018

It simplifies to $27$.

#### Explanation:

${\left(\setminus \frac{{3}^{2}}{{3}^{- 3}}\right)}^{\setminus \frac{3}{5}} = {\left({3}^{\left(2 + 3\right)}\right)}^{\setminus \frac{3}{5}} = {\left({3}^{5}\right)}^{\setminus \frac{3}{5}} = {3}^{5 \setminus \cdot \setminus \frac{3}{5}} = {3}^{3} = 27$

Mar 28, 2018

${3}^{3}$. See below

#### Explanation:

You have several ways to resolve. This could be the simpliest...

First operate whithin the braquets using laws of exponents

${3}^{2} / {3}^{- 3} = {3}^{2 - \left(- 3\right)} = {3}^{5}$

Now, apply exponent $\frac{3}{5}$ to this result

(3^5)^(3/5)=3^(5·3/5)=3^3