# How do you simplify ((5x)/(3yz))^-3 and write it using only positive exponents?

Jan 14, 2018

See a solution process below:

#### Explanation:

First, let's use this rule of exponents to rewrite the expression as:

$\frac{1}{4} y + \left(\frac{1}{4} \times \frac{3}{2}\right) \implies$

${\left(\frac{5 x}{3 y z}\right)}^{\textcolor{red}{- 3}} \implies \frac{1}{\frac{5 x}{3 y z}} ^ \textcolor{red}{- - 3} \implies \frac{1}{\frac{5 x}{3 y z}} ^ 3$

Now, use these rules of exponents to eliminate the outer exponent:

$a = {a}^{\textcolor{red}{1}}$ and ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

$\frac{1}{\frac{{5}^{\textcolor{red}{1}} {x}^{\textcolor{red}{1}}}{{3}^{\textcolor{red}{1}} {y}^{\textcolor{red}{1}} {z}^{\textcolor{red}{1}}}} ^ \textcolor{b l u e}{3} \implies \frac{1}{\frac{{5}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}} {x}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}}}{{3}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}} {y}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}} {z}^{\textcolor{red}{1} \times \textcolor{b l u e}{3}}}} \implies \frac{1}{\frac{{5}^{3} {x}^{3}}{{3}^{3} {y}^{3} {z}^{3}}} \implies \frac{1}{\frac{125 {x}^{3}}{27 {y}^{3} {z}^{3}}} \implies \frac{27 {y}^{3} {z}^{3}}{125 {x}^{3}}$